Aruldas-500-problems.pdf

QUANTUM MECHANICS
500 Problems with Solutions
G. Aruldhas
Formerly Professor and Head of Physics
and Dean, Faculty of Science
University of Kerala
PHI Learning P fcte taftM
New Delhi-110001
2011

1295.00
QUANTUM MECHANICS: 500 Problems with Solutions
G. Aruldhas
© 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be
reproduced in any form, by mimeograph, or any other means, without permission in writing from the
publisher.
ISBN-978-81-203-4069-5
The export rights of this book are vested solely with the publisher.
Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus,
New Delhi-110001 and Printed by V.K. Batra at Pearl Offset Press Private Limited,
New Delhi-110015.

To
my wife, Myrtle
Our children
Vinod & Anitha, Manoj & Bini, Ann & Suresh
and
Our grandchildren
Nithin, Cerene, Tina, Zaneta, Juana, Joshua, Tesiya, Lidiya, Ezekiel
for their unending encouragement and support

Contents
Preface
1. QUANTUM THEORY
1.1 Planck’s Quantum Hypothesis 1
1.2 Photoelectric Effect 1
1.3 Compton Effect 2
1.4 Bohr’s Theory of Hydrogen Atom 2
1.5 Wilson-Sommerfeld Quantization Rule 4
Problems 5
2. WAVE MECHANICAL CONCEPTS
2.1 Wave Nature of Particles 17
2.2 Uncertainty Principle 17
2.3 Wave Packet 18
2.4 Time-dependent Schrodinger Equation 18
2.5 Physical Interpretation of t) 18
2.5.1 Probability Interpretation 18
2.5.2 Probability Current Density 19
2.6 Time-independent Schrodinger Equation 19
Problems 21
3. GENERAL FORMALISM OF QUANTUM MECHANICS
3.1 Mathematical Preliminaries 44
3.2 Linear Operator 45
3.3 Eigenfunctions and Eigenvalues 45
3>4 Hermitian Operator 45
3.5 Postulatesjof3Quantum Mechanics 46
3.5.1 Postulate 1—Wave Function 46
3.5.2 Postulate 2—Operators 46

viii • Contents
3.5.3 Postulate 3—Expectation Value 47
3.5.4 Postulate 4— Eigenvalues 47
3.5.5 Postulate 5—Time Development of a Quantum System 47
3.6 General Uncertainty Relation 47
3.7 Dirac’s Notation 48
3.8 Equations of Motion 48
3.8.1 Schrodinger Picture 48
3.8.2 Heisenberg Picture 48
3.8.3 Momentum Representation 49
Problems 50
4. ONE-DIMENSIONAL SYSTEMS 84-125
4.1 Infinite Square Well Potential 84
4.2 Square Well Potential with Finite Walls 85
4.3 Square Potential Barrier 86
4.4 Linear Harmonic Oscillator 86
4.4.1 The Schrodinger Method 86
4.4.2 The Operator Method 86
4.5 The Free Particle 87
Problems 88
5. THREE-DIMENSIONAL ENERGY EIGENVALUE PROBLEMS 126-158
5.1 Particle Moving in a SphericallySymmetric Potential 126
5.2 System of Two InteractingParticles 127
5.3 Rigid Rotator 127
5.4 Hydrogen Atom 127
Problems 129
6. MATRIX FORMULATION AND SYMMETRY 159-175
6.1 Matrix Representation of Operators and Wave Functions 159
6.2 Unitary Transformation 159
6.3 Symmetry 160
6.3.1 Translation in Space 160
6.3.2 Translation in Time 160
6.3.3 Rotation in Space 161
6.3.4 Space Inversion 161
6.3.5 Time Reversal 162
Problems 163
7. ANGULAR MOMENTUM AND SPIN 176-214
7.1 Angular Momentum Operators 176
7.2 Angular Momentum CommutationRelations 176
7.3 Eigenvalues of J2 and J7 177

1
Contents • IX
7.4 Spin Angular Momentum 177
7.5 Addition of Angular Momenta 178
Problems 179
#8. TIME-INDEPENDENT PERTURBATION 215-247
8.1 Correction of Nondegenerate Energy Levels 215
8.2 Correction to Degenerate Energy Levels 215
Problems 217
9. VARIATION AND WKB METHODS 248-270
9.1 Variation Method 248
9.2 WKB Method 248
9.3 The Connection Formulas 249
Problems 250
#10 TIME-DEPENDENT PERTURBATION 271-286
10.1 First Order Perturbation 271
10.2 Harmonic Perturbation 272
10.3 Transition to Continuum States 272
10.4 Absorption and Emission of Radiation 273
10.5 Einstein’s A and B Coefficients 273
10.6 Selection Rules 273
Problems 274
11. IDENTICAL PARTICLES 287-307
11.1 Indistinguishable Particles 287
11.2 The Pauli Principle 287
11.3 Inclusion of Spin 288
Problems 289
12. SCATTERING 308-329
12.1 Scattering Cross-section 308
12.2 Scattering Amplitude 308
12.3 Probability Current Density 309
12.4 Partial Wave Analysis of Scattering 309
12.5 The Bom Approximation 310
Problems 311
13. RELATIVISTIC EQUATIONS 330-342
13.1 Klein-Gordon Equation 330
13.2 Dirac’s Equation for a Free Particle 330
Problems 332

X • Contents
14. CHEMICAL BONDING 343-357
14.1 Bom-Oppenheimer Approximation 343
14.2 Molecular Orbital and Valence Bond Methods 343
14.3 Hydrogen Molecule-ion 344
14.4 MO Treatment of Hydrogen Molecule 345
14.5 Diatomic Molecular Orbitals 345
Problems 347
APPENDIX 359-360
INDEX 361-363

Preface
This comprehensive, in-depth treatment of quantum mechanics in the form of problems with
solutions provides a thorough understanding of the subject and its application to various physical and
chemical problems. Learning to solve problems is the basic purpose of a course since it helps in
understanding the subject in a better way. Keeping this in mind, considerable attention is devoted to
work out these problems. Typical problems illustrating important concepts in Quantum Mechanics
have been included in all the chapters. Problems from the simple plugged-ins to increasing order of
difficulty are included to strengthen the students’ understanding of the subject.
Every effort has been taken to make the book explanatory, exhaustive, and user-friendly.
Besides helping students to build a thorough conceptual understanding of Quantum Mechanics, the
book will also be of considerable assistance to readers in developing a more positive and realistic
impression of the subject.
It is with a deep sense of gratitude and pleasure that I acknowledge my indebtedness to my
students for all the discussions and questions they have raised. I express my sincere thanks to the
Publishers, PHI Learning, for their unfailing cooperation and for the meticulous processing of the
manuscript. Finally, I acknowledge my gratitude to my wife, Myrtle, and our children for the
encouragement, cooperation, and academic environment they have provided throughout my career.
Above all, I thank my Lord Jesus Christ who has given me wisdom, knowledge, and guidance
throughout my life.
G. Aruldhas

Chapter
Quantum Theory
Quantum physics, which originated in the year 1900, spans the first quarter of the twentieth century.
At the end of this important period, Quantum Mechanics emerged as the overruling principle in
Physics.
1.1 Planck’s Quantum Hypothesis
Quantum physics originated with Max Planck’s explanation of the black body radiation curves.
Planck assumed that the atoms of the walls of the black body behave like tiny electromagnetic
oscillators, each with a characteristic frequency of oscillation. He then boldly put forth the following
suggestions:
1. An oscillator can have energies given by
En = nhv, n = 0, 1, 2, ... (1.1)
where v is the oscillator frequency and h is Planck’s constant whose value is
6.626 x 10-34 Js.
2. Oscillators can absorb energy from the cavity or emit energy into the cavity only in discrete
units called quanta, i.e.,
AEn = Anhv= hv (1.2)
Based on these postulates, Planck derived the following equation for the spectral energy
density uv of black body radiation:
8Khv* _______________________________________________dv__3.
Uy c3 exp (hv/kT) - 1
1.2 Photoelectric Effect
On the basis of quantum ideas, Einstein succeeded in explaining the photoelectric effect. He extended
Planck’s idea and suggested that light is not only absorbed or emitted in quanta but also propagates
l

2 • Quantum Mechanics: 500 Problems with Solutions
as quanta of energy h v, where v is the frequency of radiation. The individual quanta of light are
called photons. Einstein’s photoelectric equation
1 9
hv = hvf j +—mv (1.4)
explained all aspectsof photoelectric effect. In Eq. (1.4), h vis the energy of theincident photon, hv0
is thework function of the metallic surface, and v0 is the thresholdfrequency. Since the rest mass
of photon is zero,
E hv h
E = cp or p = — = — = — (1.5)
C C A
1.3 Compton Effect
Compton allowed x-rays of monochromatic wavelngth K to fall on a graphite block and measured
the intensity of scattered x-rays. In the scattered x-rays, he found two wavelengths—the original
wavelength X and another wavelength X which is larger than X. Compton showed that
h
X' - X = ----- (1 - cos 0) (1.6)
m0c
where m0 is the rest mass of electron and <
f>is the scattering angle. The factor hlmyc is called the
Compton wavelength.
1.4 Bohr’s Theory of Hydrogen Atom
Niels Bohr succeeded in explaining the observed hydrogen spectrum on the basis of the following
two postulates:
(i) An electron moves only in certain allowed circular orbits which are stationary states in the
sense that no radiation is emitted. The condition for such states is that the orbital angular
momentum of the electron is given by
mvr - nh, n = 1, 2, 3, ... (1.7)
where h = h tln is called the modified Planck’s constant, v is the velocity of the electron
in the orbit of radius r, and m is the electron mass.
(ii) Emission or absorption of radiation occurs only when the electron makes a transition from
one stationary state to another. The radiation has a definite frequency vmn given by the
condition
hvmn = Em - E n (1.8)
where Em and E„ are the energies of the states m and n, respectively.
According to Bohr’s theory, the radius of the nth orbit is
r" = ^ 7 ’ k = (L9)
where £q is the permittivity of vacuum and its experimental value is 8.854 x 10~12 C2 N_1 m-2.

Quantum Theory • 3
The
= 0.53 A
me
radius of the first orbit is called Bohr radius and is denoted by a0, i.
_*2
a0 =
In terms of a0, from Eq. (1.9), we have
r„ = nla0
The total energy of the hydrogen atom in the nth state is
1 13.6
e.
E - - me
32n 2e$h2 n
eV, n = 1, 2, 3, ...
(1.10)
(1.11)
( 1.12)
When the electron drops from the with to nth state, the frequency of the emitted line vmn is given by
, me
hv = -------------
mn 9 7 . i
32j z SqH 2 2
n m )
m > n > 1
For hydrogen-like systems,
E„ =
Z 2me4 1
32jr2£&h2 n2 ’
n = 1, 2, 3, ...
(1.13)
(1.14)
---- —
U
The parameters often used in numerical calculations include the fine structure constant a and the
Rydberg constant R given by
a =
47c£0ch 137
(1.15)
4
R = = 10967757.6 m"1 (1.16)
Sewell
The Rydberg constant for an atom with a nucleus of infinite mass is denoted by R„, which is the
same as R in (1.16).
Different spectral series of hydrogen atom can be obtained by substituting different values for
m and n in Eq. (1.13).
(i) The Lyman series
(ii) The Balmer series
(iii) The Paschen series
92
VL m )
m
m
= 2, 3, 4, ...
= 3, 4, 5, ...
(1.17)
(1.18)

(iv) The Brackett series
J = R f i r
—7 , m = 5, 6, 7,
(v) The Pfund series
m
5 m y
m = 6, 1 , 8, ...
(1.20)
(1.21)
1.5 Wilson-Sommerfeld Quantization Rule
In 1915, Wilson and Sommerfeld proposed the general quantization rule
j Pi dqt = nth, n, = 0, 1, 2, 3, ... (1.22)
where § is over one cycle of motion. The qf s and p,’s are the generalized coordinates and
generalized momenta, respectively. In circular orbits, the angular momentum L = mvr is a constant
of motion. Hence, Eq. (1.22) reduces to
n.h
mvr = —
—, n = 1, 2, 3, ..
2k
(1.23)
which is Bohr’s quantization rule. The quantum number n = 0 is left out as it would correspond to
the electron moving in a straight line through the nucleus.

PROBLEMS
1.1 The work function of barium and tungsten are 2.5 eV and 4.2 eV, respectively. Check whether
these materials are useful in a photocell, which is to be used to detect visible light.
Solution. The wavelength A. of visible light is in the range 4000-7000 A. Then,
...........i - , . , , he (6.626 x 10 34Js) (3 x 10? m/s) „ _ „
Energy of 4000 A light = — = -----------------j----------------- —
-------- = 3.106 eV
A (4000x 10 10m)(l.6 x l 0 ~19 J/eV)
. . , 6.626 x 10”34 x 3x 108 , „
Energy of 7000 A light = ---------------^ = 1.77 eV
7000 xlO “10x 1.6 xlO -19
The work function of tungsten is 4.2 eV, which is more than the energy range of visible light. Hence,
barium is the only material useful for the purpose.
1.2 Light of wavelength 2000 A falls on a metallic surface. If the work function of the surface is
4.2 eV, what is the kinetic energy of the fastest photoelectrons emitted? Also calculate the stopping
potential and the threshold wavelength for the metal.
Solution. The energy of the radiation having wavelength 2000 A is obtained as
he (6.626 x 10-34J s) (3 x 108 m/s)
A (2000 x 10“10m)(1.6 x 10“19 J/eV)
Work function = 4.2 eV
KE of fastest electron = 6.212 - 4.2 = 2'.012 eV
Stopping potential = 2.012.V
he
Threshold wavelength Aq =
= 6.212 eV
Work function
in n/n x ill ^
^0
(6.626 x 10 J s) (3 x 108 m/s) x
= ----------------------------— - 2958 A
(4.2 eV)(1.6 x 10~'y J/eV)
1.3 What is the work function of a metal if the threshold wavelength for it is 580 nm? If light of
475 nm wavelength falls on the metal, what is its stopping potential?
Solution.
„ T , ^ .. he (6.626 x 10-34J s) (3 x 108 m/s) „ , „
Work function = -r- = ---------- — ----- ------- — —
-------- - 2.14 eV
4) (580 x 10 m)(1.6 x 10"19 J/eV)
he (6.626 x 10_34Js) (3 x 108 m/s)
Energy of 475 nm radiation = —
r- = -------------- „-----------------7
7
;-------- - 2.62 ev
& (475 x 10_9m)(1.6 x 10-19 J/eV)
Stopping potential = 2.62 - 2.14 = 0.48 V
1.4 How much energy is required to remove an electron from the n = 8 state of a hydrogen atom?
-13.6 eV
Solution. Energy of the n = 8 state of hydrogen atom = ----- ------ = -0.21 eV .
8
The energy required to remove the electron from the n = 8 state is 0.21 eV.
1.5 Calculate the frequency of the radiation that just ionizes a normal hydrogen atom.
Solution. Energy of a normal hydrogen atom = -13.6 eV

Frequency of radiation that just ionizes is equal to
| , 13.fre V (1. 6 x 10--»J/eV) = 3 2 8 4 x l0 ,SH z
h 6.626 x 10“34 Js
1.6 A photon of wavelength 4 A strikes an electron at rest and is scattered at an angle of 150° to
its original direction. Find the wavelength of the photon after collision.
Solution.
AA = X - A= — (1 - cos 150°)
moc
6.626 x 1(T34Js x 1.866 ,
= 0.045 A
(9.11 x 10 kg)(3 x 10 m/s)
A + 0.045 A = 4.045 A
1.7 Whenradiation ofwavelength 1500A is incident on a photocell, electrons are emitted. If the
stoppingpotential is4.4 volts,calculate the work function, threshold frequency and threshold
wavelength.
he
Solution. Energy of the incident photon = —
A
(6.626 x 10“34 J s) (3 x 108 m/s) o „
= 8.28 ev
(1500 x 10“10m)(1.6 x 10~19 J/eV)
Work function = 8.28 - 4.4 = 3.88 eV
3.88 eV (1.6 x l0 - 19J/eV) _ _ 14 „
Threshold frequency vf, = ------------ :--------- —
---------- = 9.4 x 10 Hz
6.626 X 10“34Js
Threshold wavelength Ao = — = ^ ^ = 3191 A
v0 9.4 x 1014 s_1
1.8 If a photon has wavelength equal to the Compton wavelength of the particle, show that the
photon’s energy is equal to the rest energy of the particle.
Solution. Compton wavelength of a particle = h/m0c~
he
Wavelength of a photon having energy E = —
/ E
Equating the above two equations, we get
h he
m{)c ~ T O
T E = n w
which is the rest energy of the particle.
1.9 x-rays of wavelength 1.4 A are scattered from a block of carbon. What will be the wavelength
of scattered x-rays at (i) 180°, (ii) 90°, and (iii) 0°?
Solution.
A = A + - ^ - ( 1 - c o s 0), A = 1.4 A
m0c

h 6.626 x 10 34 Js nMA i
— rj - —0.024 A
9.1 x 1(T31kg (3 x 108m/s)
(i) A' = A + — x 2 = 1.45 A
m0c
(ii) A '= A + — = 1.42 A
m0c
(iii) A' = A + — (1 - 1) = 1.4 A
moc
1.10 Determine the maximum wavelength that hydrogen in its ground state can absorb. What
would be the next smallest wavelength that would work?
Solution. The maximum wavelength corresponds to minimum energy. Hence, transition from
n = 1 to n = 2 gives the maximum wavelength. The next wavelength the ground state can absorb is
the one for n = 1 to n = 3.
The energy of the ground state, E = -13.6 eV
__ 2 3 ^
Energy of the n = 2 state, E2 = — eV = -3.4 eV
_^
Energy of the n = 3 state, E3 = — eV = -1.5 eV
Maximum wavelength =
E2 - E 1
_ (6.626 x 10 J s) (3 x 10s m/s)
10.2 eV x 1.6x10 19 J/eV
= 122 x 10~9 m = 122 nm
he
Next maximum wavelength = —----- tt =103 nm
£3 - *1
1.11 State the equation for the energy of the nth state of the electron in the hydrogen atom and
express it in electron volts.
Solution. The energy of the nth state is
E„= -
me4 1
8e^h2 n2
-(9.11 x 10~31 kg) (1.6 x 10~19 C)4
8(8.85 x 10_12C2 N"1m“2)2(6.626 x 10“34J s ) V
-21.703 x 10“19 j _ 21.703 x 10-19 J
n2 1.6 x 10-19n2 J/eV
13.56
n
eV

1.12 Calculate the maximum wavelength that hydrogen in its ground state can absorb. What would
be the next maximum wavelength?
Solution. Maximum wavelength correspond to minimum energy. Hence the jump from ground state
to first excited state gives the maximum X.
Energy of the ground state = -13.6 eV
Energy of the first excited state = -13.6/4 = -3.4 eV
Energy of the n = 3 state = -13.6/9 = -1.5 eV
Maximum wavelength corresponds to the energy 13.6 - 3.4 = 10.2 eV
c he
Maximum wavelength = —= —----- —
_ (6.626 x 10~34J s) x (3.0 x 108m/s)
10.2 x 1.6 x lO “19J
= 122 x 10~9 m = 122 nm
The next maximum wavelength corresponds to a jump from ground state to the second excited state.
This requires an energy 13.6 eV - 1.5 eV = 12.1 eV, which corresponds to the wavelength
_ (6.626 x 10~34J s) x (3.0 x 108m/s)
12.1 x 1.6 x 10“19J
= 103 x 10~9 m = 103 nm
,1.13 A hydrogen atom in a state having binding energy of 0.85 eV makes a transition to a state
with an excitation energy of 10.2 eV. Calculate the energy of the emitted photon.
Solution. Excitation energy of a state is the energy difference between that state and the ground
state.
Excitation energy of the given state = 10.2 eV
Energy of the state having excitation energy 10.2 eV = -13.6 + 10.2 = - 3.4 eV
Energy of the emitted photon during transition from - 0.85 eV to -3.4 eV
= -0.85 - (-3.4) = 2.55 eV
Let the quantum number of -0.85 eV state ben and that of -3.4 eVstate be m. Then,
= 0.85 or n2 = 16 or n = 4
«2
13 6
— -r- = 3.4 or m2 = 4 or m = 2
m2
The transition is from n = 4 to n = 2 state.
1.14 Determine the ionization energy of the He+ ion. Also calculate the minimum frequency a
photon must have to cause ionization.
Solution. Energy of a hydrogen-like atom in the ground state = -Z2 x 13.6 eV
Ground state energy of He+ ion = - 4 x 13.6 eV = - 54.4 eV
Ionization energy of He+ ion = 54.4 eV

The minimum frequency of a photon that can cause ionization is
il J‘+.‘+ CVU.UAH1 ^
V =
* j±leyq:6xlO-'»JteV) = , 3 , 3 6 x 1 Q „ H z
h 6.626 xlO _34Js
1.15 Calculate the velocity and frequency of revolution of the electron of the Bohr hydrogen atom
in its ground state.
Solution. The necessary centripetal force is provided by the coulombic attraction, i.e.
mv2 ke1 , 1
r2 ’ * 4ne.
o
Substituting the value of r from Eq. (1.9), the velocity of the electron of a hydrogen atom in its
ground state is obtained as
Vi =
e2 (1.6 x 10 19C)2
2£0h 2(8.85 x 10' 12 C2N_1 m-2)(6.626 x 10"34 Js)
= 2.18 x 106ms-1
In r
Period T = —
n
Substituting the value of r and vj, we obtain the frequency of revolution of the electron in the ground
state as
me4 (9.11xlO “31kg)(1.6xlO 19C)4
v'i - “
4£%h3 4(8.85 x 10~12 C2N 1m"2)(6.626 x 10 34 Js)3
= 6.55 x 1015 Hz
1.16 What is the potential difference that must be applied to stop the fastest photoelectrons emitted
by a surface when electromagnetic radiation of frequency 1.5 x 1015Hz is allowed to fall on it? The
work function of the surface is 5 eV.
Solution. The energy of the photon is given by
hv = (6.626 x 10~34 Js)(1.5 x 1015 s-1)
= (6-626 x IQ"34 J s )(1 .5 x l0 15s 1) = 6 2 U
1.6 x 10-19 J/eV
Energy of the fastest electron = 6.212 - 5.0 = 1.212 eV
Thus, the potential difference required to stop the fastest electron is 1.212 V
1.17 x-rayswith A = 1.0 A are scattered from a metal block. Thescattered radiation is viewed at
90° to theincidentdirection. Evaluate the Compton shift.
Solution. The compton shift
h „ ^ (6.626 x lO “34J s ) ( l - c o s 90°)
Aa = ----- (l-cos<*>) = --------------- -----------------------—
moc (9.11 x 10 kg)(3 x 10 m s )
= 2.42 x 10~12 m = 0.024 A

1.18 From a sodium surface, light of wavelength 3125 and 3650 A causes emission of electrons
whose maximum kinetic energy is 2.128 and 1.595 eV, respectively. Estimate Planck’s constant and
the work function of sodium.
Solution. Einstein’s photoelectric equation is
he he
— = — + kinetic energy
A Aq
he he
3125 x 10~10m 4)
+ 2.128 eV x (1.6 x 10~19J/eV)
he
3650 x 10“10m 4>
hr
= + 1.595 eV (1.6 x 10~19J/eV)
1
he
lO"10 1 3125 3650
= 0.533 x 1.6 x lO-19 J
0.533 x 1.6 x 10"19 x 10“10 x 3125 x 3650 _
h= ------------------------------------ 5— ---------------- Js
525 x 3 x 108
= 6.176 x lO' 34 Js
From the first equation, the work function
he = (6.176 x 10~34 J s)(3 x IQ8 m/s) _ ^ x Lfi x 1(J_I9 ;
A) 3125 x 10-10 m
= 2.524 x 1.6 x 10-19 J = 2.524 eV
1.19 Construct the energy-level diagram for doubly ionized lithium.
Solution.
Z2 x 13.6 „ 9x1 3 .6 „
E --------------------eV = ---------- -— eV
n
122.4
n2
eV
Ei = -122.4 eV
E3 = -13.6 eV '
E2 = -30.6 eV
£4 = -7.65 eV
These energies are represented in Fig. 1.1.
E(eV)
0
-7.65
-13.6
-30.6
-122.4
Fig. 1.1 Energy level diagram for doubly ionized lithium (not to scale).

1.20 What are the potential and kinetic energies of the electron in the ground state of the hydrogen
atom?
Solution.
Potential energy =
1 e2
4t
c
£q r
Substituting the value of r from Eq. (1.9), we get
4
me
Potential energy =
-
—— - = -2E, = -27.2 eV
16n £qH
Kinetic energy = total energy - potential energy
= -13.6 eV + 27.2 eV = 13.6 eV
1.21 Show that the magnitude of the potential energy of an electron in any Bohr orbit of the
hydrogen atom is twice the magnitude of its kinetic energy in that orbit. What is the kinetic energy
of the electron in the n = 3 orbit? What is its potential energy in the n - 4 orbit?
Solution.
Radius of the Bohr orbit rn = n2a0
1 e2 1 e2 27 2
Potential energy = - - — = - - — = --— eV
4 ^ 0 rn n 0 O n 2
Kinetic energy = Total energy - Potential energy
13.6 „ 27.2 w 13.6 , T
= -----5- eV + — — eV = — — eV
n n n
13 6
KE in the n - 3 orbit = ——— = 1.51 eV
27 2
Potential energy in the n = 4 orbit = ------— = - 1.7 eV
16
1.22 Calculate the momentum of the photon of largest energy in the hydrogen spectrum. Also
evaluate the velocity of the recoiling atom when it emits this photon. The mass of the atom =
1.67 x 10-27 kg.
Solution. The photon of the largest energy in the hydrogen spectrum occurs at the Lyman series
limit, that is, when the quantum number n changes from °° to 1. For Lyman series, we have
2 2 m = 2, 3, 4, ...
For the largest energy, m = Hence,
U r
£ ' " ’ 'r-?' ! ^ ^ •
hv h
Momentum of the photon = — = — = hR
c A
= (6.626 x 10~34 Js) (1.0967 x 107 n r 1)
= 7.267 x IQ"27 kg m s~'

12 • OiiantumMechanics: 500 Problems with Solutions
momentum
Velocity of recoil of the atom = ---------------
J mass
= 7-266 x l 0 - 27kgm£ ^ ^ 4 35m s- 1
1.67 x 10-27 kg
1.23 Show that the electron in the Bohr orbits of hydrogen atom has quantized speeds v„ = coin,
where a is the fine structure constant. Use this result to evaluate the kinetic energy of hydrogen atom
in the ground state in eV.
Solution. According to the Bohr postulate,
mvr = nh, n = 1, 2, 3, ...
The coulombic attraction between the electron and the proton provides the necessary centripetal
force, i.e.,
mv2 ke2 ^ _ 1
ke2
mvr = -----
v
Combining the two equations for mvr, we obtain
ke ke
-----= nh or v = ——
v nh
ke2 c ac . ke2
v = ------- = — since a = ——
ch n n cn
.2~2
1 2 1 c a
Kinetic energy = —my = —m— j-
l n
1 (9.1 x IQ-31 kg)(3 x 108ms x)2 1
2 1372 n2
21.8179 x 10_19J 21.8179 x 10"19J
n2 n2(1.6 x 10-19 J/eV)
= 13.636- y eV
n
Kinetic energy in the ground state = 13.636 eV
1.24 In Moseley’s data, the K„ wavelengths for two elements are found at 0.8364 and 0.1798 nm.
Identify the elements.
Solution. The K„ x-ray is emitted when a vacancy in the K-shell is filled by an electron from the
L-shell. Inside the orbit of L-electron, there are z-protons and the one electron left in the K-shell.
Hence the effective charge experienced by the L-electron is approximately (Z - l)e. Consequently,
the energy of such an electron is given by
(Z - l)213.6 eV

Then, the frequency of the Ka line is
Since v = dX, we have
(Z - l)2 13.6eV ( 1 1 A
vKa - ;---------- ------ ~
h v1 2 .
3 (Z - l ) 2 13.6eV
4 h
_ 3 (Z - l)2 (13.6eV)(1.6 x 10~19J/eV)
4 6.626 x l(T34Js
= 2.463 x 1015 (Z - l)2 s”1
3 x 108ms 1 „ is 9 i
--------------------= 2.463 x 1015(Z - l)2 s_1
0.8364 x l 0 “9 m
Z - 1 = 12.06 or Z = 13
Hence the element is aluminium. For the other one
3 x 108m s-1 _ is 9 i
----------------- 5— = 2.463 x 1015(Z - l)2 s"1
0.1798 x l 0 _9m
Z - 1 = 26, Z = 27
The element is cobalt.
1.25 Using the Wilson-Sommerfeld quantization rule, show that the possible energies of a linear
harmonic oscillator are integral multiples of hv0, where v0 is the oscillator frequency.
Solution. The displacement x with time t of a harmonic oscillator of frequency v{) is given by
x = x0 sin (2nv0t) (i)
The force constant k and frequency v0 are related by the equation
V
° = i ^ O
T * = 4^ (“)
Potential energy V = ^ k x 2 = 2T^mvfixjf sin2 (2/rvy) (iii)
Kinetic energy T = ^ m i2 = 2n2m vlxl cos2 (2nv0t) (iv)
Total energy E = T + V = l ^ m VqXq (v)
According to the quantization rule,
px dx = nh or m § xdx = nh (vi)
When x completes one cycle, t changes by period T = 1/vf,. Hence, substituting the values of x and
dx, we obtain
lv 0
47t2mv%xl J cos2 (2jrv0t)dt = nh, n = 0, 1, 2, ...

2ft2mv0x0 - nh or x0 =
r , ^/2
nh
2n 2mv0 j
Substituting the value of x0 in Eq. (v), we get
En - nhv0 - rihco, n = 0, 1, 2, ...
That is, according to old quantum theory, the energies of a linear harmonic oscillator are integral
multiples of hv0 = ha).
1.26 A rigid rotator restricted to move in a plane is described by the angle coordinate 9. Show that
the momentum conjugate to 6 is an integral multiple of h. Use this result to derive an equation for
its energy.
Solution. Let the momentum conjugate to the angle coordinate be p# which is a constant of motion.
Then,
In 2n
J pg dd = Pg J dd = 27C
pg
0 0
Applying the Wilson-Sommerfeld quantization rule, we get
—
-----—
i
27tpe = nh or Jpe = nh^j n =0, 1,2, ...
Since p e = Ico, 1(0 = nh. Hence, the energy of a rotator is
U „ = 4 r - . : n = 0, 1, 2, ...
_ ... 21 ,J
1.27 -
' The lifetime of the n - 2 state of hydrogen atom is10-8 s. How many revolutionsdoes, an
electron in the n = 2 Bohr orbit make during this time?
Solution. The number of revolutions the electron makes in onesecond in the n = 2 Bohr orbit is
E2 (13.6eV)(1.6 x 10~19 J/eV)
Vl~ h ~ 4(6.626 x 10 34 Js)
= 0.821 x 1015 s_1
No. of revolutions the electron makes in 10-8 s = (0.821 x 1015 s_1)(10r8 s)
= 8.21 x 106
1.28 In a hydrogen atom, the nth orbit has a radius 10“5m.Find the value of n. Write a note on
atoms with such high quantum numbers.
Solution. In a hydrogen atom, the radius of the nth orbit rn is

Atoms having an outermost electron in an excited state with a very high principal quantum
number n are called Rydberg atoms. They have exaggerated properties. In such atoms, the valence
electron is in a large loosely bound orbit. The probability that the outer electron spends its time
outside the Z - 1 other electrons is fairly high. Consequently, Zeff is that due to Z-protons and
(Z —1) electrons, which is 1. That is, Zeff = 1 which gives an ionization energy of 13.6 eV/n2 for
all Rydberg atoms.
1.29 When an excited atom in a state £, emits a photon and comes to a state Ef , the frequency of
the emitted radiation is given by Bohr’s frequency condition. To balance the recoil of the atom, a part
of the emitted energy is used up. How does Bohr’s frequency condition get modified?
Solution. Let the energy of the emitted radiation be Ey = h v and Eie be the recoil energy. Hence,
where M is the mass of recoil atom
Substituting the value of Ete, the Bohr frequency condition takes the form
where v is the frequency of the radiation emitted and M is the mass of the recoil nucleus.
1.30 Hydrogen atom at rest in the n = 2 state makes transition to the n = 1 state.
(i) Compute the recoil kinetic energy of the atom.
(ii) What fraction of the excitation energy of the n = 2 state is carried by the recoiling atom?
Solution. Energy of the n = 2 -> n = 1 transition is given by
E, - Ef = h v + Ek
By the law of conservation of momentum,
Recoil momentum of atom = momentum of the emitted y-ray
hv
where c is the velocity of light,
= 10.2 x 1.6 x 10“19 J
(i) From Problem 1.29, the recoil energy
Ek = ------- (M-mass of the nucleus)
1 AA~2 '
2Me
(E2 - E rf
2Me2
(10.2 x 1.6 x 10“19J)2
2(9.1 x 10“31 kg) 1836(3 x 108 m/s)2
= 8.856 x lO' 27 J
= 5.535 x 10-8 eV

(ii) Excitation energy of the n = 2 state is 10.2 eV. Then,
Recoil energy 5.535 x l 0 _8eV 9
------------------—— = ----- ----------------= 5.4 X10*
Excitation energy 10.2 ev
1.31 In the lithium atom (Z = 3), the energy of the outer electron is approximated as
(Z - a )2 13.6 eV
n2
where <7is the screening constant. If the measured ionization energy is 5.39 eV, what is the value
of screening constant?
Solution. The electronic configuration for lithium is 1s2.2 s1. For the outer electron, n = 2. Since
the ionization energy is 5.39 eV, the energy of the outer electron E = -5.39 eV. Given
( Z - c r)2 13.6eV
n2
Equating the two energy relations, we get
1.32 The wavelength of the La line for an element has a wavelength of 0.3617 nm. What is the
element? Use (Z - 7.4) for the effective nuclear charge.
Solution. The La transition is from n = 3 to n = 2. The frequency of the La transition is given by
(Z - a )2 13.6eV
22
= - 5.39eV
(Z - o f
Z - £7= 1.259
<7= 3 —1.259 = 1.741
£ _ (Z - 7.4)2 13.6eV f j ____ 1_'
A ~ h v 22 32 ,
0.3617 x 10“9m
3 x 108m/s (Z - 7.4)2 (13.6eV x 1.6 x 10~19J/eV)
6.626 x 10~34 J s
5
x —
8.294 x 1017 s’1 = (Z - 7.4)2 (0.456 x 1015 s”1)
Z - 7.4 = 42.64 or Z = 50.04
The element is tin.

Chapter
Wave Mechanical Concepts
2.1 Wave Nature of Particles
Classical physics considered particles and waves as distinct entities. Quantum ideas firmly
established that radiation has both wave and particle nature. This dual nature was extended to
material particles by Louis de Broglie in 1924. The wave associated with a particle in motion, called
matter wave, has the wavelength X given by the de Broglie equation
where p is the momentum of the particle. Electron diffraction experiments conclusively proved the
dual nature of material particles in motion.
2.2 Uncertainty Principle
When waves are associated with particles, some kind of indeterminacy is bound to be present.
Heisenberg critically analyzed this and proposed the uncertainty principle:
where Ax is the uncertainty in the measurement of position and Apx is the uncertainty in the
measurement of the x-component of momentum. A more rigorous derivation leads to
Two other equally useful forms are the energy time and angular momentum-polar angle relations
given respecting by
h h
p m (2.1)
Ax ■Apx ~ h (2.2)
(2.3)
h
AE ■At > -
2 (2.4)
ALz A</>> %
r ~ 2
17
(2.5)

2.3 Wave Packet
The linear superposition principle, which is valid for wave motion, is also valid for material particles.
To describe matter waves associated with particles in motion, we requires a quantity which varies
in space and time. This quantity, called the wave function 'F(r, t), is confined to a small region in
space and is called the wave packet or wave group. Mathematically, a wave packet can be
constructed by the superposition of an infinite number of plane waves with slightly differing ^-values,
as
'P(jc, t) = JA(k) exp [ikx - ia>(k)t] dk (2.6)
where k is the wave vector and a) is the angular frequency. Since the wave packet is localized, the
limit of the integral is restricted to a small range of ^-values, say, (k0-A k )< k < (k0+ AA;). The speed
with which the component waves of the wave packet move is called the phase velocity p which is
defined as
v = — (2-7)
p k
The speed with which the envelope of the wave packet moves is called the group velocity vg given
by r
v = — (2.8)
* dk
2.4 Time-dependent Schrodinger Equation
For a detailed study of systems, Schrodinger formulated an equation of motion for 'F(r, t):
2m
V2 + V(r) ¥ ( r ,0 (2.9)
The quantity in the square brackets is called the Hamiltonian operator of the system. Schrodinger
realized that, in the new mechanics, the energy E, the momentum p, the coordinate r, and time t have
to be considered as operators operating on functions. An analysis leads to the following operators for
the different dynamical variables:
E —
» ih ^ - ,p —
>-iftV,
at
r r, (2.10)
2.5 Physical Interpretation of 'F(r, 0
2.5.1 Probability Interpretation
A universally accepted interpretation of >
F(r, t) was suggested by Bom in 1926. He interpreted 'P*'P
as the position probability density P (r, t):
|2
P(r, t) = ¥*(!■, t) ¥(r, t) = ^ (r, t) (2.11)

Wave Mechanical Concepts • 19
The quantity^ ( r , f)| dr isthe probability of finding the system at time t in the elementary volume
d t surrounding the pointr.Since the total probability is 1,we have
J |'P ( r , 0 |2 dT = l (2.12)
If ¥ is not satisfying this condition, one can multiply Y by a constant, say N, so that N*¥ satisfies
Eq. (2.12). Then,
N2 ]'¥ (r,t)1 dr =l (2.i3)
The constant N is called the normalization constant.
2.5.2 Probability Current Density
The probability current density j (r, t) is defined as
ih
j(r ’ t ) = 2iH ('FV'r * " ^ V4/) (2‘14>
It may be noted that, if Y is real, the vectorj (r, t) vanishes. The functionj (r, t) satisfies the equation
of continuity
Yt p (r,t) + V j ( r ,t) = 0 (2.15)
Equation (2.15) is a quantum mechanical probability conservation equation. That is, if the probability
of finding the system in some region increases with time, the probability of finding the system
outside decreases by the same amount.
2.6 Time-independent Schrodinger Equation
If the Hamiltonian operator does not depend on time, the variables r a
nd tof the wave function
*F(r, f) can be separated into two functions y/(r) and (pit) as
*F(r, t) = y/(r) (pit) (2.16)
Simplifying, the time-dependent Schrodinger equation, Eq. (2.9),splits into the following two
equations:
1 dtp _ iE
~ ~ Y
(p{t) dt
- A v 2 + y (,)
(2.17)
y/(r) = Ey/(r) (2.18)
The separation constant E is the energy of the system. Equation (2.18) is the time-independent
Schrodinger equation. The solution of Eq. (2.17) gives
(p(t) = Ce~iEm (2.19)
where C is a constant.

20 • Quantum Mechanics; 500 Problems with Solutions
'Ptr, f) now takes the form
*F(r, t) = y/(r)e-4Etm (2-2°)
The states for which the probability density is constant in time are called stationary states, i.e.,
P(r, t) = I'FCr, f)l2 = constant in time (2.21)
Admissibility conditions on the wave functions
(i) The wave function v
F(r, t) must be finite and single valued at every point in space.
(ii) The functions 'F and V^must be continuous, finite and single valued.

PROBLEMS
2.1 Calculate the de Broglie wavelength of an electron having a kinetic energy of 1000 eV.
Compare the result with the wavelength of x-rays having the same energy.
Solution. The kinetic energy
2
T = ~ = 1000 eV = 1.6 X 10~16 J
2m
6.626 x 10“34js
P [2 x (9.11 x 10 31 kg) x (1.6 x 10-16 J]1/2
For x-rays,
= 0.39 x 10~10m = 0.39 A
c hc
Energy = —
A =
(6.626 x 10 J s) x (3 x 108m/s)
1.6 x 10 -16
Wavelength of x-rays
= 12.42 x 10~10 m = 12.42 A
12.42 A
= 31.85
de Broglie wavelength of electron 0.39 A
2.2 Determine the de Broglie wavelength of an electron that has been accelerated through a
potential difference of (i) 100 V, (ii) 200 V.
Solution.
(i) The energy gained by the electron = 100 eV. Then,
_2
A--
IX
2m
= 100 eV = (100 eV)(1.6 x 10~19 J/eV) = 1.6 x 10“17 J
p = [ 2 (9.1 x 10“13 kg)(1.6 x 10-17 J)]1/2
= 5.396 x 10-24 kg ms
^-1
(ii)
P 5.
= 1.228 x
l _ =
2m
p = [2(9.1 :
= 7.632 x
6.626 x 10“34Js
96 x 10 24 kg ms 1
10"10 m = 1.128 A
>00 eV = 3.2 x 10-17 J
10-31 kg)(3.2 x 10-17 J)]1/2
1CT24 kg ms"1
6.626 x 10“34Js
A= - =
P 7.632 x 10“24 kg ms"1
= 0.868 x 10“10 m = 0.868 A

2.3 The electron scattering experiment gives a value of 2 x 10 15m for the radius of a nucleus.
Estimate the order of energies of electrons used for the experiment. Use relativistic expressions.
Solution. For electron scattering experiment, the de Broglie wavelength of electrons used must be
of the order of 4 x 10' 1
5m, the diameter of the atom. The kinetic energy
T = E - m%c2= yjc2p 2+ m%c4 - m0c2
('.T + mf]c2)2 = c2p 2 + m^c4
m
^c4
2
1+
m0c2 j
c2
/?2 = m$c4
2 2 , 24
c p + m0c

1+
p = M qC
*0
*- y
2
1+
m0c y
2
1+
X2m^c2
hl 2 2
IT = ^
+ 1
-1
2
1/2
1+
(6.626 x lO _34Js)
(16 x 10-30 m2) x (9.11 x 10”31 kg)2 x (3 x 108 m/s)2
= 3.6737 x 105
T = 605. lni(fi2
= 605.1 x (9.11 x 10“31 kg) x (3 x 108 m/s)2
= 496.12 x 1 0 - J = 496-12X |1
9
° '13j
1.6 x 10“ J/eV
= 310 x 106 eV = 310 MeV
2,4 Evaluate the ratio of the de Broglie wavelength of electron to that of proton when (i) both have
the same kinetic energy, and (ii) the electron kinetic energy is 1000 eV and the proton KE is
100 eV.
-31
+ 1

Solution.
(i) A =
V2"1
!7!
A
2
ini Tx = 1000 eV;
•y/2m2T2
X of electron
X of proton
T2 = 100 eV
A
x1
m
2T2
mxTx
1836 meT
m
JT
= V1836 = 42.85
X of electron (1836x 100
1000
= 13.55
X of proton
2.5 Proton beam is used to obtain information about the size and shape of atomic nuclei. If the
diameter of nuclei is of the order of 10“15m, what is the approximate kinetic energy to which protons
are to be accelerated? Use relativistic expressions.
Solution. When fast moving protons are used to investigate a nucleus, its de Broglie wavelength
must be comparable to nuclear dimensions, i.e., the de Broglie wavelength of protons must be of the
order of 10“15 m. In terms of the kinetic energy T, the relativistic momentum p is given by (refer
Problem 2.3)
p = rriQc, 1+ - 1 X - — = 10“15 m
P
/ 2
T
- 1
1 + 2"
1 c
h2 2 2
-7Y = moc
Substitution of X, m0, h and c gives
T = 9.8912 x 10“u J = 618.2 MeV
2.6 Estimate the velocity of neutrons needed for the study of neutron diffraction of crystal
structures if the interatomic spacing in the crystal is of the order of 2 A. Also estimate the kinetic
energy of the neutrons corresponding to this velocity. Mass of neutron = 1.6749 x 10 27 kg.
Solution, de Broglie wavelength
X = 2 x 10
X = —
my
6.626 x 10 34 Js
-10
m
or v =
h
mX
(1.6749 x IQ 27 kg)(2 x 10 1
Um/s)
= 1.978 x 103 ms' 1
-t, 1 2
Kinetic energy T = —my -
1 12
(1.6749 x 10 2/ kg) (1.978 x 10J m s'1)
= 3.2765 x 10“21 J = 20.478 x 10“3 eV
2.7 Estimate the energy of electrons needed for the study of electron diffraction of crystal
structures if the interatomic spacing in the crystal is of the order of 2 A.

Solution, de Broglie wavelength of electrons = 2 A = 2 x 10-10 m
p 2 (MX)2
Kinetic energy T =
T =
2m 2m
(6.626 x 10~34Js)2
v*l° rn i 1 ^ 1n~31
2 x (2 x 10 m) (9.11 x 10 kg)
= 60.24 x lO-19 J = 37.65 eV
2.8 What is the ratio of the kinetic energy of an electron to that of a proton if their de Broglie
wavelengths are equal?
Solution.
mi = mass of electron, m2 - mass of proton,
Vj = velocity of electron, v2 = velocity of proton.
, h h
A - --------- or mxVi = m2v2
m2 2
mi I ^ mivi
^ f 1 2^
= ml 2 ^ 2
Kinetic energy of electron m,
_ _ — -------------- ---------- = — = 1836
Kinetic energy of proton m1
2.9 An electron has a speed of 500 m/s with an accuracy of 0.004%. Calculate the certainty with
which we can locate the position of the electron.
Solution.
Momentum p = m = (9.11 x 10-31 kg) x (500 m/s)
Ad
— x 100 = 0.004
P
0.004(9.11 x10 31 kg) (500 m/s)
^ 100
= 182.2 x 10“34 kg m s"1
h 6.626 x l 0“34Js
Ax = — = --------------- ------------ - = 0.0364 m
AP 182.2 x 10 kgms^
The position of the electron cannot be measured to accuracy less than 0.036 m.
2.10 The average lifetime of an excited atomic state is ICr9s. If the spectral line associated with
the decay of this state is 6000 A, estimate the width of the line.
Solution.
At = 10~9 s, A = 6000 x Ip"10 m = 6 x 10~7 m

AE ■At = AA ■At = —
A1 2 An
A2 36 x KT14 m2 „ „ ,4
AA = - — — = ----------------------------------------------------------- — = 9.5 x 10 m
AncAt An (3 x 10 m/s) x (1O' 9s)
2.11 An electron in the « = 2 state of hydrogen remains thereon the average of about 10~8 s, before
making a transition to n = 1 state.
(i) Estimate the uncertainty in the energy of the n = 2 state.
(ii) What fraction of the transition energy is this?
(iii) What is the wavelength and width of this line in the spectrum of hydrogen atom?
Solution. From Eq. (2.4),
/•x h 6.626 x 10"34Js
(l) AE > ------- = ----------------------
AnAi 4tt x 10-8 s
= 0.527 x 10~26 J = 3.29 x 10"8 eV
(ii) Energy of n = 2 n = 1 transition
= -13.6 eV
' ___l_x
22 l2j
= 10.2 eV
AE 3.29 x 10-8 eV
Fraction —
— = = 3.23 x 10 9
E 10.2 eV
, he (6.626 x 10-34 Js) x (3 x 108 m/s)
' ' 1
7 ” IQ
E (10.2 x 1.6 x 10”19J)
= 1.218 x 10~7 m = 122 nm
AE AA . . AE ,
— = x or U =
AA= (3.23 x 10-9) (1.218 x 10"7 m)
= 3.93 x 10~16 m = 3.93 x 10“7 nm
2.12 An electron of rest mass m$ is accelerated by an extremely high potential of V volts. Show
that its wavelength
he
A = - 2m1/2
[eV (eV + 2m0c )]
Solution. The energy gained by the electron in the potential is Ve. The relativistic expression for
mQ
cl 2 ^
kinetic energy = -------- —_ - m0c . Equating the two and rearranging, we get
(1 - z/cz)ul
moC2 2 yj
---------------------m0c = Ve
(1 - v2/c2)1/2
(1 - v2/c2)1/2 = — ^
2
Ve + m0c2

2 2 4
j _ v m0c
c2(Ve + m0c2)2
v2 (Ve + m0c2)2 - m^c4 _ Ve(Ve + 2m0c2)
c2 (Ve + m0c2)2 (Ve + m0c2)2
c[Ve(Ve + 2m0c2) f 2
v = -
de Broglie Wavelength
Ve + m^c2
A= _h_ = h(1 - v2/c2)1/2
M V OT0 V
ft m0c2 Ve + O
T
qC
2
"Jo Ve + m0c2 c[Ve(Ve+ 2m0c2) f 2
he
[Ve (Ve + 2m0c2)]1/2
2.13 A subatomic particle produced in a nuclear collision is found to have a mass such that Me2
= 1228 MeV, with an uncertainty of ± 56 MeV. Estimate the lifetime of this state. Assuming that,
when the particle is produced in the collision, it travels with a speed of 108 m/s, how far can it travel
before it disintegrates?
Solution.
Uncertainty in energy AE = (56 X 106 eV) (1.6 X 10-19 J/eV)
_ h 1 (1.05 x 10~34 J s ) ________ _________
2 AE 2 (56 x 1.6 x 10“13J)
= 5.86 x l(r24 s
Its lifetime is about 5.86 x 10-24 s, which is in the laboratory frame.
Distance travelled before disintegration = (5.86 X 10-24 s)(108 m/s)
= 5.86 x 10~16 m
2.14 A bullet of mass 0.03 kg is moving with a velocity of 500 m's-1. The speed is measured up
to an accuracy of 0.02%. Calculate the uncertainty in x. Also comment on the result.
Solution.
Momentum p - 0.03 x 500 = 15 kg m s~*
Ap
— x 100 = 0.02
P
0.02 x 15 „ ^ , ,
Ap = — — — = 3 x 10 kg m s
h 6.626 x 10"34Js , ^
Ax ~ —
— = ----------------- -------- = 1.76x10 m
2Ap 4 x 3 x 10 km/s

As uncertainty in the position coordinate x is almost zero, it can be measured very accurately. In
other words, the particle aspect is more predominant.
2.15 Wavelength can be determined with an accuracy of 1 in 108. What is the uncertainty in the
position of a 10 A photon when its wavelength is simultaneously measured?
Solution.
AA ~ 10'8 m, A = 10 x 10“10 m = 10'9 m
h h ,
p = J or Ap = — AA
Ax x AA x h
Ax -Ap = ---------;-------
A
From Eq. (2.3), this product is equal to h/2. Hence,
(Ax) (AA) h _ h
A2 ~ An
Ax = t 4 t = 10 ^ mg = 7-95 x 1 0 12 m
AnAA4n x 10“8m
2.16 If the position of a 5 k eV electron is located within 2 A, what is the percentage uncertainty
in its momentum?
Solution.
Ax = 2 x 10“10 m; Ap ■Ax = -p-
An
= (6,626 x U T * is ) = 2 635 x 10_2J kg m s_,
AnAx 4n (2 x 10“10 m)
p = V2mT = (2 x 9.11 x 10' 31 x 5000 x 1.6 x 10“19)1/2
= 3.818 x 1(T23 kg m s' 1
Ap 2.635 x 10 25
Percentage of uncertainty = — x 100 = --------------- x 100 - 0.69
P 3.818 x l0 “23
2.17 The uncertainty in the velocity of a particle is equal to its velocity. If Ap ■Ax = h, show that
the uncertainty in its location is its de Broglie wavelength.
Solution. Given Av = v. Then,
Ap - mAv = mv = p
Ax x Ap = h or Ax •p = h
Ax = — = A
P
2.18 Normalize the wave function y/(x) = A exp (-ax2), A and a are constants, over the domain
—
O
O< X < oo.
Solution. Taking A as the normalization constant, we get
A2 J y/* ff dx = A2 J exp (-2ax2) dx = 1

Using the result (see the Appendix), we get
f exp ( - 2ax2)dx = J —
J u In
2a
* - ( t )
1/4
y/(x) = exp (-ax )
2.19 A particle constrained to move along the x-axis in the domain 0 < x < L has a wave function
yAx) = sin (nnx/L), where n is an integer. Normalize the wave function and evaluate the expectation
value of its momentum.
Solution. The normalization condition gives
L
2 nnx
N 2 J sinz dx = 1
L 1 2nnx ^
1 - cos—- — Iax = 1
N 2 4 = 1 or A
T
The normalized wave function is yj2/L sin (nnx)IL]. So,
-ih —
dx
y/dx
nrtx nTtx ,
j sin —
— cos — —dx
Lj Li q Li Li
.. tin f . 2nnx ,
= -in —
r- sin —-— dx = 0
1} J
0 L
2.20 Give the mathematical representation of a spherical wave travelling outward from a point and
evaluate its probability current density.
Solution. The mathematical representation of a spherical wave travelling outwards from a point is
given by
y/(r) = — exp (ikr)
where A is a constant and k is the wave vector. The probability current density

ih
j= v * v Y)
ih I , i2
2 ^ IAI
ih I i2
2^IAI
Ixl2
J*r
ikr
-ikr
r
v y
-ikr Jkr
r
J
( i k _ - i k r e - i k r )
^
1
1
{ i k A r
-s
V 1 )
r r
V r 2 J
r -2 ik^ hk
r J 4
2.21 The wave function of a particle of mass m moving in a potential V(x) is ^(x, t) =
( km 2 ^
A exp | -ikt — — x , where A and k are constants. Find the explicit form of the potential V(x).
Solution.
TCx, t) - A exp —
ikt —
kmx
a r
dx
d2y¥
dx2
2kmx
¥
f 2km Ak2m2x2 '
K' ~ + h2
!)T/
m — = km>
dt
Substituting these values in the time dependendent Schrodinger equation, we have
kh ='
2m
2km 4k2m2x2
— + ------------ + V(x)
kh= kh - 2mk1x1 + V(x)
V(x) = 2mk2x2
2.22 The time-independent wave function of a system is yAx) —A exp (ikx), where £ is a constant.
Check whether it is normalizable in the domain < x < oo. Calculate the probability current density
for this function.
Solution. Substitution of y/(x) in the normalization condition gives
|7V|2 J y/2 dx = |w|2 J dx = 1
As this integral is not finite, the given wave function is not normalizable in the usual sense. The
probability current density

j = <*Vw* - V)
= ^ - A 2 [eikx(-ik)e~ikx - e ~ ikx(ik)eikx]
2m 1 1
2m 11 m 1 1
2.23 Show that the phase velocity p for a particle with rest mass w0 is always greater than the
velocity of light and that p is a function of wavelength.
Solution.
(O h
Phase velocity p = — = vA A = —
k P
Combining the two, we get
pvp = hv= E = (c2p2 + r^c 4)112
P*p = C
P 1 +
Nl/2
= cp 1 +
2 2
m0c
P2 j
vp = c
1+
™
p
2c2
p2 /
l/2
or p > c
1 +
mQC2A2
ft2
Hence vp is a function of A.
2.24 Show that the wavelength of a particle of rest mass m^ with kintic energy T given by the
relativistic formula
A =
he
yjr2 + 2m0c2T
Solution. For a relativistic particle, we have
Now, since
E2 = c2p 2 + m^c4
E = T + rtiff
(T + m0c2)2 = c p2 + mfcc4
T2 + 2m0c2T + m£c4 = c2p 2 + m^c4
cp = yfr2 + 2m0c2T
de Broglie wavelength A = — =
he
4-Tl + 2m0c T

2.25 An electron moves with a constant velocity 1.1 x 106 m/s. If the velocity is measured to a
precision of 0.1 per cent, what is the maximum precision with which its position could be
simultaneously measured?
Solution. The momentum of the electron is given by
p = (9.1 x 10“31 kg) (1.1 x 106 m/s)
= 1 x 10-24 kg m/s
Av _ Ap _ 0.1
v p 100
Ap = p x 10 3 = 10 z/ kg m/s
,-27
Ax
h 6.626 x 10"34 Js
= 6.6 x 10“' m
4nAp n x 10 27 kg m/s
2.26 Calculate the probability current density j(x) for the wave function.
y/(x) - u{x) exp [i<f>(x),
where u, <
pare real.
Solution.
y/(x) = u(x) exp dtp); y/*(x) - u(x) exp (-it,V)
dw du . dtp
exp (itp) + ih — exp (itp)
ox ox ox
dyf* du . dtp
— — = — exp ( - 10) - i u - ^ exp (itp)
j(x) =
ih
2m ¥
dx dx
d y ^ _ dy_
dx
I//*
dx
ih
2m
ih
2m
ih
2m
du _i$ . d(j) _
— e * - iu-z— e
dx dx
du
dx
id , • dtp
e v A- i u — —
dx
du
dx
-2iu'
du
■2 d<
P
IU -----U-z:—
O X OX
dtp
dx
2 d<
P
dx
h
- u 2^ -
m dx
2.27 The time-independent wave function of a particle of mass m moving in a potential V(x) - a 2x2
is
yAx) = exp
m a2 7
— T x
2h
, a being a constant.
Find the energy of the system.
S o lu tio n We have

dy/
dx
2ma
x exp'
i
m a 2
---- T x
2h
d2yf
dx2
2m a
1 -
2m a
exp
m a
I 2n2
Substituting these in the time-independent Schrodinger equation and dropping the exponential term,
we obtain
2m
+ a2x2 = E
$vx2 + t£ x2 = E
E =
h a
4 2m
2.28 For a particle of mass m, Schrodinger initially arrived at the wave equation
1 92y g2y m2c2
c2 dt2 dx2 ft2
Show that a plane wave solution of this equation is consistent with the relativistic energy momentum
relationship.
Solution. For plane waves,
4'Cx, t) = A exp [/ (kx - mt)]
Substituting this solution in the given wave equation, we obtain
-C
O 2„2
= - r
m c
Multiplying by c2h2 and writing ha>= E and kh = p, we get
E2 = c2p2 + m2c4
which is the relativistic energy-momentum relationship.
2.29 Using the time-independent Schrodinger equation, find the potential V(x) and energy E for
which the wave function '
y/(x) =
f
X
x0
-X /X n
where n, x0 are constants, is an eigenfunction. Assume that V(x) —
>0 as x

Solution. Differentiating the wave function with respect to x, we get
d y
dx
n--1
-x/xn
d2y _ n ( n - 1)
dx2 r 2
x0
f n~2
X
1
/
X
x0
—
jc
/jc
q 2n
xl
1
'JL
r2
x0 l xoJ
1
2
y(x)
-jr/jtn
/ A
" '1
~X
/X
n +
-An
Substituting in the Schrodinger equation, we get
2m
n(n —1) 2n 1
x2 + xl
y + V y = E y
which gives the operator equation
E - V(x) = - —
2m
n(n —1) 2n 1
x2 + xl
E =
2m
2mxl
n(n —1) 2n
x2
2.30 Find that the form of the potential, for which y(r) is constant, is a solution of the Schrodinger
equation. What happens to probability current density in such a case?
Solution. Since y(r) is constant,
V2jif= 0.
Hence the Schrodinger equation reduces to
V y = E y or V = E
The potential is of the form V which is a constant. Since y(r) is constant, V y = V y = 0.
Consequently, the probability current density is zero.
2.31 Obtain the form of the equation of continuity for probability if the potential in the Schrodinger
equation is of the form V(r) = V^r) + iV2(r), where V and V2 are real.
Solution. The probability density P(r, t) = ¥ . Then,

The Schrodinger equation with the given potential is given by
ih
d ¥ _ - T
dt 2m
V2xV + (Vj + iV2) ¥
a y 3'?* .
Substituting the values of ih and ih , we have
ih
aP
dt
hl
2m
ih^ = -^ -[V ('PV'F* -'¥ * ¥ '¥ )+ 2iV2P]
at 2m
dt
ih
2m h
2.32 For a one-dimensional wave function of the form
'Vix, t) = A exp [itp (x, f)]
show that the probability current density can be written as
j = - A f ¥
m ' ' dxi
Solution. The probability current density j(r, t) is given by
ih
j(r, f)= — ('PV'P* - T V *)
x
P(x, t) = A exp [itj>(x, t)}
vF*(x, t) = A* exp [-10 (x, f)]
VT = ^ - = iAe'f
O X ox
V¥* =
dx dx
Substituting these values, we get
ih
J = 2m
ih
2m
Ae1
* -iA*e~i*
dtj)
dx
A e~ iAe* ^
OX
-'■
w
2--w2
l!K w 2
!?
dx
2.33 Let y/o(x) and y/(x) be the normalized ground and first excited state energy eigenfunctions of
a linear harmonic oscillator. At some instants of time, Ay/G+ By/j, where A and B are constants, is
the wave function of the oscillator. Show that (x) is in general different from zero.

Solution. The normalization condition gives
<(A% + B y ) | (Ay0 + B y )) = 1
A2(Wo I Vo) + B2{y i I V) = 1 or A2 + S2 = 1
Generally,the constants A and B are not zero. The average value of x is givenby
<x> = ((Ay0 + B y )x (A i//0 + B y ))
= A2{y/0x y0) + B2{yx x y x) + 2AB{y0 |x| y{)
since A and B are realand (% |x| y x) = ( y |x| y0). As the integrands involved is odd,
(VoxVo) = (Vi 1*1 Vi) = 0
<x> = 2AB(y/0 |x| y/)
which is not equal to zero.
2.34 (i) The waves on the surface of water travel with a phase velocity vp - ^gA /2n, where g is
the acceleration due to gravity and X is the wavelength of the wave. Show that the group velocity
of a wave packet comprised of these waves is Vp/2. (ii) For a relativistic particle, show that the
velocity of the particle and the group velocity of the corresponding wave packet are the same.
Solution.
(i) The phase velocity
v/>=
where k is the wave vector.
By definition, p = calk, and hence
The group velocity
= = I 1 = 1 l
V* dk 2 k 2
. da> dE
(n) Group velocity v„ = ——= ——
s dk dp
For relativistic particle, E2 = c2p2 + m^c4, and therefore,
dE c2p _ c2w0v-^/l - v2/c2 _
v„ =
dP Em 0 C2 y] 1 - v2/c2
2.35 Show that, if a particle is in a stationary state at a given time, it will always remain in a
stationary state.
Solution. Let the particle be in the stationary state <
F(x, 0) with energy E. Then we have
/m x , 0) = E'F(x, 0)

where H is the Hamiltonian of the particle which is assumed to be real. At a later time, let the wave
function be *F(x, t), i.e.,
Yix, t) = T(x, 0) e~iE,m
At time t,
HY(x, t) = HY(x, 0) e~lE,m
= EV(x, 0) e~,Etm
= £T(x, t)
Thus, 'FCx, t) is a stationary state which is the required result.
2.36 Find the condition at which de Broglie wavelength equals the Compton wavelength
Solution.
h
Compton wavelength Ac = -----
W
qC
where m{) is the rest mass of electron and c is the velocity of light
h
de Broglie wave length A = ----
mv
where m is the mass of electron when its velocity is v. Since
mo
m =
A =
Vi - v2/c2
h j l - v2/c2
m0v
h 1
/c2/v2 - 1
m0cv
m0cv
— ^ c 2/v2 - 1
mnc
= A . £ - i
When A = Aq,
£ _ _ 1 = 1 or ^ - 1 = 1
°2 C
— = 2 or v = —
j=
v2 V2
2.37 The wave function of a one-dimensional system is
y/(x) = Axf'e~x/a, A, a and n are constants
If y/(x) is an eigenfunction of the Schrodinger equation, find the condition on V(x) for the energy
eigenvalue E - ~h2/(2ma2). Also find the value of V(x).

Solution.
y/(x) = Axne-X
la
= Anxn le xla - - x ne x,a
dx a
d yz
dx2
= Ae -xJa
n(n - l)xn -2 2n
With these values, the Schrodinger equation takes the form
- — Ae~xla
2m
n (n - )xn~2 ~ — xn~x + V{x)Axne~xla = E A xne
n -x la
2m
n ( n - 1) 2n 1
x2 ax + a2
= E - V(x)
From this equation, it is obvious that for the energy E — -h 2/2ma2, V(x) must tend to zero as
x —
>oo. Then,
V(x) =
hl
2m
2ma2 2m
n ( n - 1) 2n
n ( n - 1) 2n 1
x2 a a2
ax
2.38 An electron has a de Broglie wavelength of 1.5 x 10"'2 m. Find its (i) kinetic energy and
(ii) group and phase velocities of its matter waves.
Solution.
(i) The total energy E of the electron is given by
E = J,c2p 2 +m
gC
4
Kinetic energy T = E - mgc2 = ■
Jc2p 2 + “■
2'-4 -— 2
m0c —m0c
he
de Broglie wavelength X = — or cp = .
P A
(6.626 x 10~34Js) (3 x 10s ms-1)
cp =
1.5 x 10-12 m
= 13.252 x 1(T14 J
2-
E0 = rrtffi = (9.1 x 10 kg) (3 X 10s m s *)
= 8.19 x 10~14 J
T= 7(13.252)2 + (8.19)2 x 10~14 J - 8.19 x 10~14 J
= 7.389 x 10~14 J = 4.62 x 105 eV

(ii) E = 7(13.252f + (8.19)2 X 10' 14 J = 15.579 x 1(T14 J
E =
V l - v 2/c2
V =
1/2
c =
1/2
1 -
8.19
15.579
(3 x 108 m s-1)
= 0.851c
The group velocity will be the same as the particle velocity. Hence,
vg = 0.851c
Phase velocity v„ = = 1.175c
v 0.851
2.39 The position of an electron is measured with an accuracy of 10-6 m. Find the uncertainty in
the electron’s position after 1 s. Comment on the result.
Solution. When t - 0, the uncertainty in the electron’s momentum is
h
Since p = mv, Ap = m Av. Hence,
Ap >
Av>
2Ax
h
2mAx
The uncertainty in the position of the electron at time t cannot be more than
ht
(Ax), = tAv >
2mAx
(1.054 xlO~34Js) Is
2(9.1 x 10“31 kg) 10-6 m
57.9 m
The original wave packet has spread out to a much wider one. A large range of wave numbers
must have been present to produce the narrow original wave group. The phase velocity of the
component waves has varied with the wave number.
2.40 If the total energy of a moving particle greatly exceeds its rest energy, show that its de Broglie
wavelength is nearly the same as the'wavelength of a photon with the same total energy.
Solution. Let the total energy be E. Then,
E2 = c2p2 + m^c4 = c2p 2
p = 7
h he
de Broglie wavelength X = — = —

For a photon having the same energy,
. he
or X = —
E
which is the required result.
2.41 From scattering experiments, it is found that the nuclear diameter is of the order of 10“15 m.
The energy of an electron in yS-decay experiment is of the order of a few MeV. Use these data and
the uncertainty principle to show that the electron is not a constituent of the nucleus.
This is very large compared to the energy of the electron in /9-decay. Thus, electron is not a
constituent of the nucleus.
<
■
__
2.42 An electron microscope operates with a beam of electrons, each of which has an energy
60 keV. What is the smallest size that such a device could resolve? What must be the energy of each
The smallest size an elecron microscope can resolve is of the order of the de Broglie wavelength of
electron. Hence the smallest size that can be resolved is 5.01 X 1 0 12 m.
The de Broglie wavelength of the neutron must be of the order of 5.01 x 10"12 m. Hence, the
momentum of the neutron must be the same as that of electron. Then,
Solution. If an electron exists inside the nucleus, the uncertainty in its position Ax = 10"15m. From
the uncertainty principle,
(10~15 m) Ap > |
The momentum of the electron p must at least be of this order.
p = 5.25 x 10-20 kgms-1
When the energy of the electron is very large compared to its rest energy,
E= cp = (3 x 108 ms_1)(5.25 x 10“20 kg m s-1)
= ------------ rj-------= 9.84 x 107 eV
1.6 x 10"19 J/eV
= 98.4 MeV
Momentum of neutron = 13.216 x 10~23 kg in s_1

2
Energy = (M is mass of neutron)
(13.216 x IQ-23 kgms- 1)2 _ 18
2 x 1836(9.1 xlO -31 kg)
5.227 x 10~18J
1.6 x 10"19J/eV
= 32.67 eV
2.43 What is the minimum energy needed for a photon to turn into an electron-positron pair?
Calculate how long a virtual electron-positron pair can exist.
Solution. The Mass of an electron-positron pair is 2mec2. Hence the minimum energy needed to
make an electron-positron pair is 2 m f 1, i.e., this much of energy needs to be borrowed to make the
electron-positron pair. By the uncertainty relation, the minimum time for which this can happen is
h
At
2 x 2m x 2
1.05 x 10“34 Js
4(9.1 x 10~31kg) (3 x 108m/s)2
/ = 3.3 x 10-22 s
which is the length of time for which such a pair exists.
2.44 A pair of virtual particles is created for a short time. During the time of their existence, a
distance of 0.35/m is covered with a speed very close to the speed of light. What is the value of mc2
(in eV) for each of the virtual particle?
Solution. According to Problem 2.43, the pair exists for a time At given by
At =
4mc2
The time of existence is also given by
. 0.35 x 10-15 m t .n_u
At = --------------------= 1.167 x 10 s
3 x 10 m/s
Equating the two expressions for At, we get
h
. = 1.167 x lO' 24 s
4 mc
2 1.05 x 10 34 Js T
me = ---------------------t— = 2.249 x 10 1
1 J
4 x 1.167 X 10 s
2.249 xlO _11J «
= 140.56 x 106 eV
1.6 x 10~19J/eV
= 140.56 MeV

2.45 The uncertainty in energy of a state is responsible for the natural line width of spectral lines.
Substantiate.
Solution. The equation
(A £ ) (A r )> | (i)
implies that the energy of a state cannot be measured exactly unless an infinite amount of time is
available for the measurement. If an atom is in an excited state, it does not remain there indefinitely,
but makes a transition to a lower state. We can take the mean time for decay t, called the lifetime,
as a measure of the time available to determine the energy. Hence the uncertainty in time is of the
order of T. For transitions to the ground state, which has a definite energy E0 because of its finite
lifetime, the spread in wavelength can be calculated from
E - E 0 =
|AE|
he
T
he |AA |
~ x 2
AA AE
A ~ E - E 0 (ii)
Using Eq. (i) and identifying At = r, we get
AA h
A ~ 2t ( E - E0) (iii)
The energy width h/r is often referred to as the natural line width.
2.46 Consider the electron in the hydrogen atom. Using (Ax),(Ap) - h, show that the radius of the
electron orbit in the ground state is equal to the Bohr radius.
Solution. The energy of the electron in the hydrogen atom is the given by
B - J t * k . 1
2m r ’ 4ke,
;o
where p is the momentum of the electron. For the order of magnitude of the position uncertainty, if
we take Ax = r, then
fi
Ap = — or (Ap)2
Taking the order of momentum p as equal to the uncertainty in momentum, we get
(Ap)2 = (p2) =
r
Hence, the total energy
E = h k£l
2mr.2

For E to be minimum, (dE/dr) = 0. Then,
dE__
dr
h2
1
^
1
4
mr3 r2
h2
kme*'
= a0
which is the required result.
2.47 Consider a particle described by the wave function ¥(*, t) = e,(kx~ eot).
(i) Is this wave function an eigenfunction corresponding to any dynamical variable or
variables? If so, name them.
(ii) Does this represent a ground state?
(iii) Obtain the probability current density of this function.
Solution.
(i) Allowing the momentum operator -ih (dJdx) to operate on the function, we have
-ih — ei(kx~ = ih(ik) ei(kx-
dx
= hk ei<
kx- m
Hence, the given function is an eigenfunction of the momentum operator. Allowing the
energy operator -ih (d/dt) to operate on the function, we have
ih — eKkx~m) = ih(-io)) eKkx'
dt
= h(aeKkx- M
)
Hence, the given function is also an eigenfunction of the energy operator with an
eigenvalue ho).
(ii) Energy of a bound state is negative. Here, the energy eigenvalue is ha, which is positive.
Hence, the function does not represent a bound state.
(iii) The probability current density -
ih"
J = 2^ {y/V ¥ * ~ V *V ¥)
2m m
2.48 Show that the average kinetic energy of a particle of mass m with a wave function y/(x) can
be written in the form
h
2
„ K r
T = —
— f
2m J
Solution. The average kinetic energy
dy/ 2
dx
dx

Integrating by parts, we obtain
c n - f
2m
yr*
dy/
dx
+
h2 7 dyr* dy/
J dx
2m dx dx
As the wave function and derivatives are finite, the integrated term vanishes, and so
<7> = £ r J
2m
dyr
dx
dx
2.49 The energy eigenvalue and the corresponding eigenfunction for a particle of mass m in a
one-dimensional potential V(x) are
A
W(x) = -j,----- 2
x + a
E = 0,
Deduce the potential V(x).
Solution. The Schrodinger equation for the particle with energy eigenvalue E = 0 is
fi2 d2y/ A
dy/
dx1
2Ax
(.x2 + a2)2
d2y/
dx2
= -2 A
4x
(x2 + a2)2 (x2 + a2)
.2x3
2A(3x2 - a 2)
(x2 + a2)3
Substituting the value of d2y/ldx1, we get
h2 2A(3x2 - a2) V(x)A
2m (x2 + a2)3 x2 + a2
V(x)
h2(3x2 - a 2)
m (x2 + a2)2

Chapter
General Formalism of
Quantum Mechanics
In this chapter, we provide an approach to a systematic the mathematical formalism of quantum
mechanics along with a set of basic postulates.
3.1 Mathematical Preliminaries
(i) The scalar product of two functions F(x) and G(x) defined in the interval a < x < b , denoted
as (F, G), is
b
(F, G) = J F*(x)G(x)dx (3.1)
a
(ii) The functions are orthogonal if
b
(F, G) = I F*(x)G(x)dx = 0 (3.2)
a
(iii) The norm of a function N is defined as
1/2
N = (F, F)m = F (x)2 dx (3.3)
a
(iv) A function is normalized if the norm is unity, i.e.,
b
(F, F) = J F*(x)F(x) dx = 1 (3.4)
a
(v) Two functions are orthonormal if
(Ft, Fj) = Sy, i,j= 1 , 2 ,3 , .. . (3.5)
44

General Formalism of Quantum Mechanics • 45
where Sy is the Kronecker delta defined by
(3.6)
(vi) A set of functions Fx(x), F2(x), ... is linearly dependent if a relation of the type
I crfCx) = 0 (3.7)
exists, where c,’s are constants. Otherwise, they are linearly independent.
3.2 Linear Operator
An operator can be defined as the rule by which a different function is obtained from any given
function. An operator A is said to be linear if it satisfies the relation
A [cJiix) + c2f 2(x)] = CjA/j(x) + c2Af2(x) (3.8)
Thecommutator of operators A and B, denoted by [A, B], is defined as
[A, B] = AB - BA (3.9)
It follows that
[A, B] = -[B, A] (3.10)
If [A, B] = 0, Aand B aresaid to commute. If AB + BA = 0, A and Bare said to anticommute. The
inverse operator A~l is defined by the relation
where a is a constant with respect to x. The function i//(x) is called the eigenfunction of the operator
A corresponding to the eigenvalue a. If a given eigenvalue is associated with a large number of
eigenfunctions, the eigenvalue is said to be degenerate.
3.4 Hermitian Operator
Consider two arbitrary functions jfm(x) and ffn{x). An operator A is said to be Hermitian if
AA"1 = A_1A = I (3.11)
3.3 Eigenfunctions and Eigenvalues
Often, an operator A operating on a function multiplies the function by a consant, i.e.,
Ay/(x) = ca//{x) (3.12)
(3.13)
An operator A is said to be anti-Hermitian if
(3.14)

Two important theorems regarding Hermitian operators are:
(i) The eigenvalues of Hermitian operators are real.
(ii) The eigenfunctions of a Hermitian operator that belong to different eigenvalues are
orthogonal.
3.5 Postulates of Quantum Mechanics
There are different ways of stating the basic postulates of quantum mechanics, but the following
formulation seems to be satisfactory.
3.5.1 Postulate 1—Wave Function
The state of a system having n degrees of freedom can be completely specified by a function 'P of
coordinates qh q2, ■
■
■
, qn and time 1 which is called the wave function or state function or state
vector of the system. X
P, and its derivatives must be continuous, finite and single valued over the
domain of the variables of V
P.
The representation in which the wave function is a function of coordinates and time is called
the coordinate representation. In the momentum representation, the wave function is a function
of momentum components and time.
3.5.2 Postulate 2—Operators
To every observable physical quantity, there corresponds a Hermitian operator or matrix. The
operators are selected according to the rule
[Q, R] = ih{q, r] (3.15)
where Q and R are the operators selected for the dynamical variables q and r, [Q, R] is the
commutator of Q with R, and {q, r] is the Poisson bracket of q and r. /
Some of the important classical observables and the corresponding operators are given in
Table 3.1.
Table 3.1 Important Observables and Their Operators
Observable Classical form Operator
Coordinates x, y, z x, y, z
Momentum P -ihV
Energy E
dt
Time t t
Kintetic energy £ _
2m 2m
Hamiltonian H
ti2 9
- T ~ V 2 + V(r)
2m

Genera] Formalism of Quantum Mechanics • 47
3.5.3 Postulate 3—Expectation Value
When a system is in a state described by the wave function Y, the expectation value of any
observable a whose operator is A is given by
{ a )= '¥ * A Y d t (3.16)
3.5.4 Postulate 4— Eigenvalues
The possible values which a measurement of an observable whose operator is A can give are the
eigenvalues a, of the equation
A'Fi = a,'F„ i = l , 2, ..., n (3.17)
The eigenfunctions form a complete set of n independent functions.
3.5.5 Postulate 5—Time Development of a Quantum System
The time development of a quantum system can be described by the evolution of state function in
time by the time dependent Schrodinger equation
a y
! dt
where H is the Hamiltonian operator of the system which is independent of time.
3.6 General Uncertainty Relation
The uncertainty (AA) in a dynamical variable A is defined as the root mean square deviation from
the mean. Here, mean implies expectation value. So,
(AA)2 = <(A - (A))2) = (A2) - (A)2 (3.19)
Now, consider two Hermitian operators, A and B. Let their commutator be
[A, B] = iC (3.20)
The general uncertainty relation is given by
(AA)(AB ) > ^ - (3.21)
In the case of the variables x and px,[x, px] - ih and, therefore,
(.Ax)(APx) > | (3.22)

3.7 Dirac’s Notation
To denote a state vector, Dirac introducted the symbol | ), called the ket vector or, simply, ket.
Different states such as ffa(r), y/b(r), ... are denoted by the kets |a), b), ... Corresponding to every
vector, |a), is defined as a conjugate vector |a)*, for which Dirac used the notation (a|, called a bra
vector or simply bra. In this notation, the functions y/a and y/b are orthogonal if
(ab) = 0 (3.23)
3.8 Equations of Motion
The equation of motion allows the determination of a system at a time from the known state at a
particular time.
3.8.1 Schrodinger Picture
In this representation, the state vector changes with time but the operator remains constant. The state
vector |y/s(t)) changes with time as follows:
= n ¥ s( 0> (3.24)
Integration of this equation gives
y,(t)) = e-iHtm¥sm (3-25)
The time derivative of the expectation value of the operator is given by
= + ^ (3.26)
3.8.2 Heisenberg Picture
The operator changes with time while the state vector remains constant in this picture. The state
vector |y/H) and operator AH are defined by
y/H) = e‘HM m ) (3.27)
An(t) = eiHmA / Hm (3.28)
From Eqs. (3.27) and (3.25), it is obvious that
¥ h ) = l^(0)> (3.29)
The time derivative of the operator AH is

3.8.3 Momentum Representation
In the momentum representation, the state function of a system t) is taken as a function of the
momentum and time. The momentum p is represented by the operator p itself and the posistion
coordinate is represented by the operator «Wp, where V, is the gradient in the p-space. The equation
or motion in the momentum representation is
2m + V(r)
For a one-dimensional system, the Fourier representation *F(x, t) is given by
¥ 0 , t) = -= L | 4>(k,t) exp(ikx)dk
V 2x
1 °°
^(k, t) = —
j= r j '¥(x,t)exp (-ikx)dk
•42k
Changing the variable from k to p, we get
The probability density in the momentum representation is |<&(p, t)|2.
(3.31)
(3.32)
(3.33)
(3.34)
(3.35)

PROBLEMS
3.1 A and B are two operators defined by Ay/{x) = y/ix) + x and BifAx) —{dlfridx) + 2 y/{x). Check
for their linearity.
Solution. An operator O is said to be linear if
O [cj/iCx) + c2f 2(x) = cxOfx{x) + c20 f 2(x)
For the operator A,
A [ci/i(x) + c2f 2(x)] = [c]/i(x) + c2f 2(x)] + x
LHS = c,A/i(x) + c2A f2(x) = Cf(x) + c2/ 2(x) + c:x + c2x
which is not equal to the RHS. Hence, the operator A is not linear.
B [cifi(x) + c2/ 2(x)] - ~ Lt'i/iW + c2/ 2(x)] + 2[cfi(x) + c2/ 2(x)]
= cx—
—f(x) + c2A f 2(x) + 2c1
/ 1(x) + 2c2/ 2(x)
dx ax
= cif(x) + 2c]fi(x) + — c2f 2(x) + 2c2/ 2(x)
= c fiM x ) + c2Bf2(x)
Thus, the operator B is linear.
3.2 Prove that the operators i(d/dx) and d2/dx2 are Hermitian.
Solution. Consider the integral J ^*1 *'-£:] Wn dx- Integrating it by parts and remembering that
y/m and y/n are zero at the end points, we get
,d_
dx
J V * i £ ) Wn dx = i ty * VnF~ - i J dx
which is the condition for i(d/dx) to be Hermitian. Therefore, id/dx is Hermitian.
] ¥ * ^ - d x =
dx
¥,n
dVn
dx - J
dWn d¥ l
dx dx
■dx
dx Wn ♦ j r
ax dx‘
Thus, d2/dx2 is Hermitian. The integrated terms in the above equations are zero since y/m and y/„ are
zero at the end points.

3.3 If A and B are Hermitian operators, show that (i) (AB + BA) is Hermitian, and (ii) (AB - BA)
is non-Hermitian.
Solution.
(i) Since A and B are Hermitian, we have
JK A P n dx = J A*K Vn dx; J ¥ *Bvndx = JB* yr* y/n dx
JV%(AB + BA) y/n dx = j yr* ABy/ndx + J yr*BAy/ndx
= J B*A* yr%y/n d x + A*B* yr* yrn dx
= J (AB + BA)*yf*y/n dx
Hence, AB + BA is Hermitian.
(ii) J W*(AB - BA) y/n dx = J (B*A* - A*B*)y/* y/n dx
= - J (AB - BA)* y/* yrn dx
Thus, AB - BA is non-Hermitian.
3.4 If operators A and B are Hermitian, show that i [A, 6] is Hermitian. What relation must exist
between operators A and B in order that AB is Hermitian?
Solution.
J y/fi [A, B] y/n dx = i j yrn
*AByn d x - i f y'*BAy/n dx
= i J B*A* y/*y/n d x - i j A*B* yr* yrn dx
= ](i[A: B}y/m)*y/n dx
Hence, i [A, B] is Hermitian.
For the product AB to be Hermitian, it is necessary that
j yr*AByrndx = J A*B* yr* yrn dx
Since A and B are Hermitian, this equation reduces to
J B*A* yr* yrn dx = j A*B* yr* yrn dx
which is possible only if B*A*yr* = A*B*yr*. Hence,
AB = BA
That is, for AB to be Hermitian, A must commute with B.
3.5 Prove the following commutation relations:
(i) [[A, B], C] + [[B, C], A] + [[C, A], B] = 0.

«A, f t q ♦ [I* c . A, ♦ nc. a ,, B - CA.8 , C- C £ * ♦ V. C] A - A [*, q
= ABC - BAC - CAB + CBA + BCA - CBA - ABC
+ ACB + C A B - ACB - BCA + BAC = 0
' 3 32 " r d d2 32 3 ]
(ii) dx ’ dx2
¥ = Kdx dx2 dx2 d*J
¥
Kdx3 dx3 j
y/ = 0
(iii) T " ’ F{X)
OX
Thus,
I/
/+ 7
A - i A = — V
d x ¥ + F dx F dx d x W
dF
SF
dx
3 6 Show that the cartesian linear momentum components (Pl, p* pi) and[ the
i i t t r .^ obev the commutation relations (l)[Lk>Pi - m p m,
components of angular momentum <L1( L3) ooey me unuu
(ii)[I*, Pk = 0, where k, I, m are the cyclic permutations of 1, 2, 3.
Solution.
k I m
(i) Angular momentum L - rk r, rm
Pk Pi Pm
I 3 _3_
Lk = rlPm - rmPl = -ift| r, ^ rm ^
[Lh pi W= - n ri
= - ft2
d dy/ d2yr dy/ d d ^ _ ^ ¥ _
r^ 3 l d n * rm dr2
Hence, [Lk, pi = ihpm-
J d 3
(ii) Lh pkW = ~ h b a T ~ r'” aT
3 fc
2 3
— y/ + ft 3 —
drk Y drk
r, —---- r — 1^ = 0
r
'3rm m3 ^
= -ft2 r,
d d f d dyi_ _ d ^ w _ , r
r‘ d ^ drk rmdr,drk 1 drk drm mdrk drt
= 0

3.7 Show that (i) Operators having common set of eigenfunctions commute; (ii) commuting
operators have common set of eigenfunctions.
Solution.
(i) Consider the operators A and B with the common set of eigenfunctions y/h i = 1, 2, 3, ...
i.e., By/i is an eigenfunction of A with the same eigenvalue a,. If A has only nondegenerate
eigenvalues, By/, can differ from y/t only by a multiplicative constant, say, b. Then,
By/i = bty/i
i.e., y/i is a simultaneous eigenfunction of both A and B.
3.8 State the relation connecting the Poisson bracket of two dynamical variables and the value of
the commutator of the corresponding operators. Obtain the value of the commutator [x, px] and the
Heisenberg’s equation of motion of a dynamical variable which has no explicit dependence on time.
Solution. Consider the dynamical variables q and r. Let their operators in quantum mechanics be
Q and R. Let {q, r} be the Poisson bracket of the dynamical variables q and r. The relation
connecting the Poisson bracket and the commutator of the corresponding operators is
as
Ay/i = aM , By/i =
Then,
ABy/t = Abjy/i = a,b,y/t
BAy/f = Bay/i = albly/t
Since ABy/i = BAy/i, A commutes with B.
(ii) The eigenvalue equation for A is
Ay/i = a ,^ , i = 1, 2, 3, ...
Operating both sides from left by B, we get
BAy/j = afiy/i
Since B commutes with A,
ABy/i = aft Vi
[Q, /?] = ih {q, r}
The Poisson bracket {;t, px) - 1. Hence,
[x, px] = ih
The equation of motion of a dynamical variable q in the Poisson bracket is
(ii)
(i)
(iii)
Using Eq. (i), in terms of the operator Q, Eq. (iii) becomes
(iv)
which is Heisenberg’s equation of motion for the operator Q in quantum mechanics.

3.9 Prove the following commutation relations (i) [Lk, r2] = 0, (ii) [Lk, p2] = 0, where r is the radius
vector, p is the linear momentum, and k, I, m are the cyclic permutations of 1, 2, 3.
Solution.
(i) [Lh r2] = [Lk, r2 + r2 + r2] .
= [Lk, rk ] + [Lk, r2] + [Lk, r2]
= rk^f-,k' rk "
*
■tLk, rk]rk +riLk>r{ + .Lkiri]ri+ rm.Lk, rm + [Lk, rm]rm
= 0 + 0 + nihrm + ihrmrt - rmihri - ihrtrm = 0
(ii) [Lk, p2] = [Lk, p] + [Lk, pf ] + [Lk, p2 ]
= p k [ L k , p k ] + [ L k , p k ]p k + P i [ , P i ] + .L k>P i ] P i + P m [ E k,
p m ] + [ L k,p m ]p m
= 0 + 0 + ihpipm + ihpmpi - ihpmpi - ihptpm = 0
3.10 Prove the following commutation relations:
(i) [x, px] = [y, py] = [z, pz = ih
(ii) [x, y]= [y, z] = [z, x] = 0
(iii) [px, py] = [py, pz] = [pz, px] = 0
Solution.
(i) Consider the commutator [x, px]. Replacing x and px by the corresponding operators and
allowing the commutator to operate on the function ip(x), we obtain
x , - i h —
dx
iff(x) = -ihx^j- + mdW
dx dx
= -ihx^f- + ihlff + itix^f-
dx dx
= ihyr
Hence,
Similarly,
x, - ih
dx
= [x,px] = ih
[y, py] = lz, pz] = ih
(ii) Since the operators representing coordinates are the coordinates themselves,
[x, y] = tv, z] = [z, x] = 0
(iii) px, py] yKx, y) = - i h — , - ih
dx dy
i/r(x,y)
= - h2
dx dy dy dx
yr(x,y)
The right-hand side is zero as the order of differentiation can be changed. Hence the
required result.

3.11 Prove the following:
(i) If y/1 and y/2 are the eigenfunctions of the operator A with the same eigenvalue, q y/x+ c2y/2
is also an eigenfunction of A with the same eigenvalue, where c, and c2 are constants.
(ii) If j/x and y/2 are the eigenfunctions of the operator A with distinct eigenvalues, then cxffx
+ c2y/2 is not an eigenfunction of the operator A, ct and c2 being constants.
Solution.
(i) We have
A V = aYb Ay/2 = al y/2
McW + c2t/f2) = Acjj/j + Ac2y/2
= ai (cl¥l + c2y/2)
Hence, the required result.
(ii) Ay/l = a l y/u and Ay/2 = a2y/2
A(c|i//x + c2y/2) = Acxy/x +m Ac2yr2
= + a2c2y/2
Thus, c^y/j + c2f/2 is not an eigenfunction of the operator A.
3.12 For the angular momentum components Lx and Ly, check whether LxLy + LyLx is Hermitian.
Solution. Since i (d/dx) is Hermitian (Problem 3.2), i (d/dy) and i (d/dz) are Hermitian. Hence Lx
and Ly are Hermitian. Since Lx and Ly are Hermitian,
J Vm (LxLy + LyLx)V n dx = J (L*L* + L*L*)y/*y/n dx
= j ( LxLy + LyLx)*V%V„dx
Thus, LxLy + LyLx is Hermitian.
3.13 Check whether the operator - ihx (d/dx) is Hermitian.
Solution.
Hence the givenoperator isnot Hermitian.
3.14 If x and p x are the coordinate and momentum operators, prove that [x, p x ] = n ih p x ~l .
Solution.
[x, px ] = [x, px~lpx = [x, px] px"-[ + px [X, pn
x-ll
= ihp?-1 + px ([x, px] p n~2 + px [x, p n
f 2)
= 2 ih p " ~ l + p x
2([x, p x p " - 3 + p x [x,
= 3 ihpZ ”1 + p 3 [x, p ”~3]
Continuing, we have [*, p x ] = n ih p x~x

3.15 Show that the cartesian coordinates (rh r2, r3) and thecartesian components of angular
momentum (Lh L^, Lg) obey the commutation relations.
(i) [Lh r{] = ihrm
(ii) [Lh rk] = 0, where k, I, m are cyclic permutations of1, 2, 3.
Solution.
(i) [Lh r{y/= (Lkr, - r,L^)i//= -ih
( d d ) ( a a >
**rm rmdrt/
¥
= -ih
= ihrmys
2dyr ^ ¥ _ _ r2 ^¥_ + rr ^
' drt J
Hence, [Lh r,] = ihrm.
(ii) [Lh rk]yf = -ih
n drm rmdrt
¥ = 0
. Thus, [Lk, rk = 0.
3.16 Show that the commutator [x, [x, H]] = -h 2lm, where H is the Hamiltonian operator.
Solution.
Hamiltonian H =
(P 2
X + Py + P z )
2m
Since
we have
[x, P y ] = [X, pz] = o, [x, px] = ih
[ x , H ] = ^ [x, P2
X] = ^ P x i x , Px] + [x,px]px )
= b ' hhp- ^ p'
[x, [x, H]] = x, ifiPx
m
3.17 Prove the following commutation relations in the momentum representation:
(i) [x, px] = [y, py] = [z, Pz] = ih
(ii) [x, y] = [y, z] = [z, x] = 0
Solution.
(i) [x,px] f ( Px) =
[x, Px] = ih
Similarly, [y, py = [z, pz] = ih

(ii) [x, y] A p x, py) = (ih)2
= - h l
d d
9PX ’ dPy
d d
f ( P x ’ Py)
d d
f ( P x > Py) = 0
dpx dpy dpy dpx
since the order of differentiation can be changed. Hence, [jt, y - 0. Similarly, [y, z] = [z, x] = 0.
3.18 Evaluate the commutator (i) [x, px], and (ii) [xyz, px].
Solution.
(i) [x, p 2] = [x, px]px + px [x, px]
= ihpx + ihpx = 2ihpx
= 2ih —
ih
dx
. d
= 2h2 —
dx
(ii) [jryz, px] = [xyz, px]px + px [xyz, px]
= xy [z, px] px + [xy, px] zpx + pxxy [z, px] + px [xy, px] z
Since [z, px], the first and third terms on the right-hand side are zero. So,
[xyz, pi] = x[y, px] zpx + [x, px] yzpx + pxx[y, px]z + px [x, px] yz
The first and third terms on the right-hand side are zero since [>’, px] = 0. Hence,
[xyz, px] = ihyzpx + ihp^z = 2ihyzpx
where we have used the result
dx
[yzfix)] y z - ^ A x )
Substituting the operator for px, we get
[xyz, px] = 2hzyZ
dx
3.19 Find the value of the operator products
(i)
(ii)
dx
+ x
d N
d ^ + X ,
U* J
+ X
— X
Solution.
(i) Allowing the product to operate on j{x), we have
d f

Dropping the arbitrary function f(x), we get
(ii)
d x
d_
d x
+ x
J
r
+ x
d x
d_
d x
+ x
d x 2
+ ' l x —-----h X + 1
d x
- x
V = T x + x
df_
d x
x f
d x z d x
d x
• + x
d x d x
x 2 - l
3.20 By what factors do the operators (x 2p 2
x + p 2
xx 2) and l/2 ( x p x + p xx ) 2 differ?
Solution. Allowing the operators to operate on the function /, we obtain
{x2p 1
x + p 2
xx 2) f = - h 2
2 d2f d2(x2f )
dx2 dx2
= -W -x
H
2..2 92/ h2 3 d (x2f )
dx2 dx dx
2J2 d2f d { , J2 df
dx1 dx
2x f + x
dx
= -h*
= - h 1
j # L + 2 f + 2 x % . + I? U + 1 I M-
dx2 Sx Sx2 Sx
2
2x2 — —+ 4 x ^~ + 2
dx*
d_
dx
f
1 ih
- ( x p x + p x x ) 2 f = - y ( * P * + Pxx )
ih
= - y (XP X + P XX)
9/ d(xf)
X dx dx
24 x * f
x ^ _ ( 2 x^ - ) + x ^ + —
3*1 dx J dx dx 2x2 ¥ -
ox dx
2x2 ? £ + 2x ¥ + x f + 2x 2 ? { + 4 x f + x i f
dx2 dx dx dx2 dx dx
&x^~ + 2jc2 ~ y + f
dx dx2
l 2 a2 . a i
2x —y + 4* 3—+ —
dx2 dx 2
= - h
{ dxL
The two operators differ by a term -(3/2)h 2
f

3.21 The Laplace transform operator L is defined by Lfix) = J e sxf(x ) dx
o
(i) Is the operator L linear?
(ii) Evaluate Le“* if s > a.
Solution.
(i) Consider the function/(x) = c-J{x) + c2f 2(x), where cj and c2 are constants. Then,
oo
L[ci/,(x) + c2f 2(x)] = j e~sx[c,/,(x) + c2f 2(x)]dx
0
= ct J e~sxf i x ) dx + c2 J e~sxf z(x)dx
o o
= CLf(x) + c2Lf2(x)
Thus, the Laplace transform operator L is linear.
“ “ -~ (s-a )x 1 °° ,
(ii) Leax = f e sxeaxdx = f e~(s~a)x dx = ---------- = — —
o o _ ( i _ a ) Jo
3.22 The operator is defined by
A - . A2 A3
e ~1+ a + 1a + ^ t + -
Show that e° = Tu where D = (d/dx) and Tx is defined by Tx
f(x) =fix + 1)
Solution. In the expanded form,
D ,d I d 2 I d 3
e = 1 + ^ + T T T T + ^T T T + - 0)
dx 2 ! dx2 V. dx3
= f(x ) + f ( x ) + ~ f ' x ) + l / " '( x ) + ... (ii)
where the primes indicate differentiation. We now have
Tifljc) = /(x + 1) (iii)
Expanding f ix + 1) by Taylor series, we get
/(* + 1) a f(x ) + f x ) + A f " (x) + ... (iy)
From Eqs. (i), (iii) and (iv), we can write
eDAx) = Tlf{x) or eD = Tx
3.23 If an operator A is Hermitian, show that the operator B = iA is anti-Hennitian. How about the
operator B = -iA?
Solution. When A is Hermitian,
Jy/*Ay/ dr = j iAy/)* y dr
For the operator B = iA, consider the integral

J y/*By/ dr = J y/*iAyr dr
i J y/*Ay/ d t = i^ A*y/*y/ dr
= - J (iAyr)* y/dt = - J (By/)* y/ dr
Hence, B = iA is anti-Hermitian. When B = -iA,
| y/*By/ dz = -i^A *yr*yr d t
= J (iA)* y/* y/ dr
Thus, B = -iA is Hermitian.
3.24 Find the eigenvalues and eigenfunctions of the operator dldx.
Solution. The eigenvalue-eigenfunction equation is
where k is the eigenvalue and yKx) is the eigenfunction. This equation can be rewritten as
where c and k are constants. If k is a real positive quantity, yf is not an acceptable function since it
tends to oo or -°o as x -» °° or When k is purely imaginary, say ia,
The function yr will be finite for all real values of a. Hence, y = ce** is the eigenfunction of the
operator d/dx with eigenvalues k = ia, where a is real.
3.25 Find the Hamiltonian operator of a charged particle in an electromagnetic field described by
the vector potential A and the scalar potential <
j>
.
Solution. The classical Hamiltonian of a charged particle in an electromagnetic field is given by
Replacing p by its operator -ihV and allowing the resulting operator equation to operate on function
f(r), we obtain
¥
Integrating In yf = kx + In c, we get
kx

m r ) - J - U v - U -ih V ---- A
c
f( r ) + e<pf(r)
2m
-ihV - —A -ih V f ~ ~ A f | + e<pf
1
2m
-h2V2f + ~ V ( A f ) + — AV/ + ~ jA 2f
c n -L
ieh
c
+ e<pf
1
2m
k2V2/ + — (V-A )f + — A -V / + — A V f + ^ - A zf
ieh ieh ieh
+ e0f
h ieh ieh
— V2 + -^— V-A + — A •V +
2m 2me me 2me
-A + /
Hence, the operatoi>representing the Hamiltonian is
h2 ieh „ . ieh e2
H = - ~ V 2 + - — V - A + A-V + -— - A + e
<
j)
2m
ieh
2me 2me
3.26 Th^ wavefunction of a particle in a state is N exp (- x2/2d), where N - (l/7ta)y4. Evaluate
(Ax) (Ap).
Solution. For evaluating (Ax) (Ap), we require the values of (x), (x2), (p) and ip2). Since iff is
symmetrical about x = 0, (x) = 0. Now,
(x2) = N 2 j x 2 exp
(p) = - //iA^2 J exp
- x
a
v y
d x
dx
a
exp
2a
v
d x
= constant J x exp
A -x2A
a

d x
= 0 since the integral is odd.
(p2) = (-ih)2 N2 J exp
J exp
-x
2a
v j dx
exp
f 2 A
-X
2a
v
d x
h2N 2
a a
v /
d x
h2N 2
a
J x2 exp
( -
>
-x
a
v
d x
h2 h2 h2
a 2a 2a

Refer the Appendix. Also,
(Ax)2 (Ap)2 = <
X
2} (p2) =
(Ax) (Ap) = -
3.27 Show that the Unear momentum is not quantized.
Solution. The operator for the ^-component of linear momentum is -ih (d/dx). Let y/k(x) be its
eigenfunction corresponding to the eigenvalue ak. The eigenvalue equation is
~ih~(L ^ k^X) = ^
dyfk(x)
y/k(x) h
T ak dx
Integrating, we get
yrk(x) = C exp j - akx
where C is a constant. The function f/k(x) will be finite for all real values of ak. Hence, all real values
of ak are proper eigenvalues and they form a continuous spectrum. In other words, the linear
momentum is not quantized.
3.28 Can we measure the kinetic and potential energies of a particle simultaneously with arbitrary
precision?
Solution. The operator for kinetic energy, T = -(h2/2m) V2. The Operator for potential energy,
V = V(r). Hence,
- y - V 2,V
2m
¥ , . — v v ¥ ) V2
2m ¥
h ■
>
Since the operators of the two observables do not commute, simultaneous measurement of both is
not possible. Simultaneous measurement is possible if V is constant or linear in coordinates.
3.29 If the wave function for a system is an eigenfunction of the operator associated with the
observable A, show that (A") = (A)n.
Solution. Let the eigenfunctions and eigenvalues of the operator A associated with the observable
A be ^and a, respectively. Then,
(A") = J y/*An¥ dr = Jy/*A"~lAy/ dr
= a J y/*An~ly/ dr = a 2jy*A "~V dr
= a nf y/*y/ dr = a n

(A") = (J W * A ¥ d r f = (a J ¥* ¥ d T )n = a n
Thus, <A"> = (A)".
3.30 The wave function y/ of a system is expressed as a linear combination of normalized
eigenfunctions $, i = 1, 2, 3, ... of the operator a of the observable A as j/= X c;$- Show that
Solution.
(A") = X k ,|2 a”, cupi = arfi, i= 1, 2, 3, ...
i
¥ = X c«$> c«= J <!>?¥*dr, i = 1, 2, 3, ...
I -o
o
(A") = J y/*any/d r = X X ct cs / <
Ptan<t>
j dr
i j -<*>
= E X ct cjaj J QTtj dr = X k ,l2 a,"
since the <
/>
’s are orthogonal.
3.31 The Hamiltonian operator of a system is H = -(d2/dx2) + x2. Show that Nx exp (-x1/!) is an
eigenfunction of H and determine the eigenvalue. Also evaluate N by normalization of the function.
Solution.
y/ = Nx exp (-x?/2), N being a constant
Hyr =
r A1
d 2
— r + *
dx
Nx exp
( 2 ^
X
'~2
v /
_TT
. r *2 i d r *2 i 2
( i
X
= Nx exp
2 dx
exp
2
- x exp
“ T
V . V V -
= 3Nx exp = 3j^
Hence, the eigenvalue of H is 3. The normalization condition gives
N 2 J x2e~x dx = 1
N 2 = 1 (refer the Appendix)
N =
Jn
The normalized function y/ =
'_ 2j
1/2 ( X
x exp
~~2
J

3.32 If A is a Hermitian operator and y/ is its eigenfunction, show that (i) (A2) = J IAy/1
2 d t and
(ii) (A2) > 0.
Solution.
(i) Let the eigenvalue equation for the operator be
Ay/= ay/
Let us assume that y/ is normalized and a is real. Since the operator A is Hermitian,
(A2) = J y/'*A2y/ dr = j A* y/*Ay/ d t
= J |Ay/2 dr
(ii) Replacing A y /by ay/, we get
(A2) = ]ay/2 dT = a 2y/?dr
= a2y/2 d t = a
2
> 0
3.33 Find the eigenfunctions and nature of eigenvalues of the operator
d2 | 2 d
dx2 x dx
Solution. Let y/ be the eigenfunction corresponding to the eigenvalue A. Then the eigenvalue
equation is given by
/ ■
y
d2 | 2 d
Kdx2 x dx
y/ = Xy/
Consider the function u = xy/. Differentiating with respect to x, we get
du dy/
dx W + X -d^
d2u _ d y / dy/ d2y/ ^ dy/ d2y/
— + “I" X — it " ^ X
dx dx dx dx dx dx2
Dividing throughout by x, we obtain
1 d2u
X dx2
2 d_ df_
x dx dx2
¥
Combining this equation with the first of the above two equations, we have
1 d2u d2u
,= A y / or ^ = Xu
x dx2 J-2
The solution of this equation is
where cx and c2 are constants.
m= c,e + c-,e
dx
-VJjt

For u to be a physically acceptable function, VA must be imaginary, say, Also, at x = 0, u = 0.
Hence, Cj + c2 = 0, c( = -c2. Consequently,
u = Ci (e'P* - e ‘P
x), y/= —ci (e'^x - e '&
*)
sm Bx
w - c ---------
x
3.34 (i) Prove that the function y/ = sin (kx) sin (k2y) sin (k$z) is an eigenfunction of the Laplacian
operator and determine the eigenvalue, (ii) Show that the function exp (ik ■r ) is simultaneously an
eigenfunction of the operators -ihV and ~h2V2 and find the eigenvalues.
Solution.
(i) The eigenvalue equation is
V > =
92 d2 d2 '
dx2 + dy2 + dz2
sin k]X sin k^y sin k3z
= - (ki + k2 + k3 ) sin kx
x sin k2y sin k3z
Hence, y/ is an eigenfunction of the Laplacian operator with the eigenvalue -(k + k + k).
(ii) -ihVe‘(kr}= hkeikr
-h 2V2e'<kr) = +h2k2e,(kr]
That is, exp (ik ■
r) is a simultaneous eigenfunction of the operators -ihV and -h2V2, with
eigenvalues hk and h2k2, respectively.
3.35 Obtain the form of the wave function for which the uncertainty product (Ax) (Ap) = h/2.
Solution. Consider the Hermitian operators A and B obeying the relation
[A, B] = iC (i)
For an operator R, we have (refer Problem 3.30)
J|/?(H2 r f r > 0 (ii)
Then, for the operator A + imB, m being an arbitrary real number,
J(A - imB)* yr*(A + imB) y/ d t > 0 (iii)
Since A and B are Hermitian, Eq. (iii) becomes
J yr*(A - imB) (A + imB) yr d t > 0
J yr*(A2 - mC + m2B2)yr d t > 0
(A2>- m(C) + m2(B2>> 0 (iv)
The value of m, for which the LHS of Eq. (iv) is minimum, is when the derivative on the LHS with
respect to m is zero, i.e.,
0 = -(C) + 2m (B2) or m = (v)
2(B)

When the LHS of (iv) is minimum,
Since
Eq. (vi) becomes
(A + itnB) yr = 0
[A - (A), B - <B>] = [A, B] = iC
[(A - (A)) + im (B - {B))]y/= 0
Identifying x with A and p with B, we get
[(jc - (x» + im {p - <p»] yr= 0, m -
2(4P)
Substituting the value of m and repalcing p by -ih{d!dx), we obtain
d y
dx
diff
¥
2(Ap f
(x - (x)) -
i(p )
h
y = 0
h2 n
dx
Integrating and replacing Ap by ti/2(Ax), we have
+ !<£>i + 1„ a,
hz
y/= N exp
h
(.x - { x ))2 , i (p)x
-----------r----h — :----
4(Axy
Normalization of the wave function is straightforward, which gives
¥ =
1
l/4
^T tiA xY
3.36 (i) Consider the wave function
exp
(x - (x))2 + i(p)x
4(Ax)2 h
y/{x) = A exp exp (ikx)
(vi)
(vii)
where A is a real constant: (i) Find the value of A; (ii) calculate (p) for this wave function.
Solution.
(i) The normalization condition gives

(ii) (P) = v * - m — y d x
= (-ih)A2 J exp
= i-iti)
f *2 ] e ,kx ( ~ + ik exp
2 2
V « J v a I a )
e~ikx dx
r 2
) /
exp
( ~ 2
- 2 x
2 2
V a '-oo I a J
xdx + (~ih)(ik) A2 J exp
—
2x
dx
In the first term, the integrand is odd and the integral is from to °°. Hence the integral vanishes.
(p) = hk (refer the appendix)
-2 x
since A2 J exp dx = 1.
3.37 The normalized wave function of a particle is y/{x) = A exp (iax - ibt), where A, a and bare
constants. Evaluate the uncertainty in its momentum.
Solution.
ifK
x) = Ae‘(a
x- b
t)
(Ap)2= (p2
) - (p)2
(p) = -ihJy/* — yfdx = ha Jy*yf dx = ha
(p2
)= -h2Jy/* iffdx
dx
= -h2A2fe~K
a
x
-b
,) ei(ax~b
,) dx
J dx2
- -h2(ia)2J yr*yfdx = h2a2
(Ap)2 = (p2
) - (p)2= h2
a2- h2
a2= 0
(Ap) = 0
3c38 Two normalized degenerate eigenfunctions y/(x) and y
f2
(x) of an observable satisfy the
condition J y/*y/2dx = a, where a is real. Find a normalized linear combination of y
rKand y
r2
,
which is orthogonal to yf - ifo
.
Solution. Let the linear combination of y
f and y
f2be
y/-cx
y/ + c2y
/2(cj, c2 are real constants)
J (c,y/i + c2yr2)* (cj^i + c2y/2)dx = 1
ci + c2 + 2qc2a = 1

As the combination y/ is orthogonal to - y/2,
J (¥ ~ ¥ 2)* (c¥i + c2y/2)dx = 0
q - c2 + c2a - cxa = 0
(ci - c2)(l - a) = 0 or C= c2
With this condition, the earlier condition on ci and c2 takes the form
1
L2 T c 2 T ^ L2
Then, the required linear combination is
c? + c? + 2c?a = 1 or c2
^J2 ~+
~2~
a
¥ =
¥ + ¥ i
y]2 + 2a
3.39 The ground state wave function of a particle of mass m is given by yKx) = exp (-a2x4/4), with
energy eigenvalue h2a 2/m. What is the potential in which the particle moves?
Solution. The Schrodinger equation of the system is given by
. ! L £ -
2m dx2
+ V e ~a2x4/4 _ ^ _ ^ _ e - a 2x4/4
m
2 J2 . -a * x* /4
2m
(-3a x + a x ) e + Ve
- a 2x4/4h 2a 2 —
cc2x414
m
h2 4 6 3 h222 h2a 2
V = —— e rr" - a l xl + -------
2m 2 2m m
3.40 An operator A contains time as a parameter. Using time-dependent Schrodinger equation for
the Hamiltonian H, show that
Solution. The ket |y/s{tj) varies in accordance with the time-dependent Schrodinger equation
ihj-t yss(t)) = H y s{t)) (i)
As the Hamiltonian H is independent of time, Eq. (3.24) can be integrated to give
Iy/s{t)) = exp (~iHt/h) y/s(Q)) (ii)
Here, the operator exp (-iHt/h) is defined by
(iii)
Equation (ii) reveals that the operator exp (-iHt/h) changes the ket | ^(0)) into ket |yss(t)). Since H
is Hermitian and t is real, this operator is unitary and the norm of the ket remains unchanged. The
Hermitian adjoint of Eq. (i) is
' iHt iH t' " 1
exp
< n ,
_ V
1
n=0 V h J n

whose solution is
= ( Ws(0)Iexp
r iHt's
v * /
(iv)
(v)
Next we consider the time derivative of expectation value of the operator As. The time
derivative of (As) is given by
(vi)
where As is the operator representing the observable A. Replacing the factors — ¥ s(t)) and
— (^(f)l and using Eqs. (i) and (iv), we get
i t {As) = Jh < ^(t)A sH - HAsWs(t)) + {¥ s(t)
dA,
dt
V jt))
(v
ii)
3.41 A particle is constrained in a potential V(x) = 0 for 0 < x < a and V(x) = °o otherwise. In the
x-representation, the wave function of the particle is given by
. . [2 . 2n x
W(x) = . — sin ------
Va a
Determine the momentum function 0(p).
Solution. From Eq. (3.35),
®(p) = .J-— J ¥ (x) exp
j2nh
In the present case, this equation can be reduced to
1 _
ipx
dx
yfnha
where
sin
2nx
dx
Integrating by parts, we obtain
h . 2nx
-----sin -------
ip a
( - ipx/h) e(-ipx,K) 2£ cos 2nx_
dx

70 • Qiinntiim Mechanics: 500 Problems with Solutions
Since the integrated term is zero,
/ =
27th 27tx(
cos------ 1 1 1 fl ff: n ' e(-ipM) I 2n s
ipa a lP J■
v ;■
''■
» ip j J K a )
. x ,
sin dx
a
2ith
ipa ^ ip
4x 2tir
a p
1 -
/ =
2 2
a P J
je(-ipm _
ap2
1]
2k ah
a2p2- 4nlh
.2 * 2
[e(-ipalh) _ j]
With this value of /,
®(p) =
2nah
[e'
(-ipalh) _
Jjtha a2p2- 4jr2h2
[e(-,pa/h) _ JJ
1]
2KV2aV2hm
a2p2 - 4n2h2
3.42 A particle is in a state | jh = (1/*)1/4 exp (-^12). Find Ax and Ap, Hence evaluate the
uncertainty product (Ax) (Apx).
Solution. For the wave function, we have
/ - 1/2 o
o - ■
W = - J x e x2dx = 0
since the integrand is an odd function of x. Now,
l/2 ~ , l/2
& = ' '
1 Y'‘ ~
r 7 -J- , „[ 1 J * _ 1
Jt
J x2e X
‘dx = 2 [ I ^ (see Appendix)

(Px) =
n
1/2
J exp
' X2 ' / -^2 d1
( 2
X
2
V /
(-ih) — - exp
dx ~~2
/
dx
x2e x dx
l / 2
h27C
m - | !
Jt
1/2 ^ 1/2 ^2
ft2 —y— = — (see Appendix)
(APx)2 = </>*> - (Px)2 = - y
The uncertainty product
(Ax)(Apx) = -
3.43 For a one-dimensional bound particle, show that
(i) ~r f 'P*(*> t) Vfx, t)dx = 0 , need not be a stationary state.
/it *'
dt
(ii) If the particle is in a stationary state at a given time, then it will always remain in a
stationary state.
Solution.
(i) Consider the Schrodinger equation and its complex conjugate form:
h2 d2
3*F(x, t)
in----r-----
dt dx7
+ V(x) ¥(*, t)
9'P*(x, t)
- i n ---- -------
dt
~ - f l + V i x )
2m fa 2
¥*(jc, t)
Multiplying the first equation by 4/* and the second by ¥ from LHS and subtracting the second from
the first, we have
ih
d y
dt dt
_ _ r _
2m
d '
^2m
dx2
d2y *
dx2
dx
dY d T * A
U/* 1 _ TJU
dx dx
Integrating over x, we get

72 • Quantum Mechanics: 500 Problems wife Solutions
Since the state is bound, Y = 0 as x -> ±°°. Hence, the RHS of the above equation is zero. The
integrated quantity will be a function of time only. Therefore,
dt
J *P*(x, t)dx = 0
(ii) Let the particle be in a stationary state at t = 0, H be its Hamiltonian which is time
independent, and E be its energy eigenvalue. Then,
H ¥(x, 0) = E'Vix, 0)
Using Eq. (3.25), we have
^(x.O = exp -
iHt
'P(^O )
Operating from left by H and using the commutability of H with exp (-iHt/h), we have
, iHt
H Vix, t) = exp| — — H'V (x,0)
- ^ - | ' P ( x , 0) = £'P(x,r)
= E exp
Thus, 'F(x, t) represents a stationary state at all times.
3.44 The solution of the Schrodinger equation for a free particle of mass m in one dimension is
Y(x, t). At t = 0,
^(x, 0) = A exp
- x
Find the probability amplitude in momentum space at t = 0 and at time t.
Solution.
(i) From Eq. (3.35),
1
4 lizh
A
yjlTCh
A
dx
exp
x _ lEL
*
dx
4 ln h _
J exp cos
px
dx
Here, the other term having sin (px/h) reduces to zero since the integrand is odd. Using the standard
integral, we get
Aa
®(p, 0) = -7==- exp
J m
.2 2
p a
4h2

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